A cold front moved in last weekend. In 8 hours overnight, the temperature outside dropped from 14 degrees to -10. What was the average temperature change for each hour?
I need the answer as quick as I can and I put it at max points!
Answer:
Step-by-step explanation:
The difference between 14 and 0 is 14, and the difference between 0 and -10 is 10. 14+10=24 for total change. For the average over 8 hours, we have 24/8=3 degrees
The number of significant figures in 0.01500 is
The number of significant figures in 0.01500 is 4.
What are significant figures?Significant figures (or significant digits) are the number of digits in a given value or a measurement, necessary to decide the accuracy and precision of measurement.
The given decimal number is 0.01500.
Certain rules help us determine the number of significant figures. These rules are as follows:
(1) All non-zero digits are significant.
(2) All zeros in between non-zero digits are significant.
(3) Zeros on the right of a decimal point and before (or to the left of) the first non-zero digit are not significant. They only represent the position of the decimal point.
(4) Zeros on the right of a decimal point are significant, provided there is no non-zero digit after them.
(5) Zeros on the right of the last non-zero digit after a decimal point are significant. So, final zeros or trailing zeros in the decimal part are significant.
(6) In a measurement value, zeros that occur on the right of the last non-zero digit are significant.
In the given decimal 0.01500, two non zero digits and two leading zeros are significant
So, number of significant figures are 4
Therefore, the number of significant figures in 0.01500 is 4.
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can someone help me please
The Pyramid of Giza is one of the largest pyramid structures still standing in Egypt. It is a right pyramid with a square base, a base length of 230 m, and height of 150 m. The area of the base is __________ The volume is ________
Answer:- The area of base is 52900 square meters. The volume is 2654000 cubic meters.
Explanation:-
Base length of right pyramid (square) a =230m
Height of right pyramid = 150m
Area of base of right pyramid[tex]=a^2=(230)^2=52900\ m^2[/tex]
Volume of right pyramid with square base[tex]=\frac{1}{3}a^2h\\=\frac{1}{3}\times52900\times150=2645000\ m^3[/tex]
Thus, the area of base is 52900 square meters and the volume is 2654000 cubic meters.
The area of the base of the pyramid is 52900 m²
The volume of the square base pyramid of Giza is 2645000 m³
The pyramid is a square base pyramid. Therefore,
volume of a square base pyramid;v = 1 / 3 Bhwhere
B = base area
h = height
h = 150 m
Therefore,
area of the base = l²
where
l = length of side
area of the base = 230² = 52900 m²
Volume = 1 / 3 × 52900 × 150
volume = 7935000 / 3
volume = 2645000 m³
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a bell tolls every 10 minutes. another bell tolls every 15 minutes. both bells toll at 6:00pm. they will toll together at what time
Both the bells will ring after 30 minutes at 6:30 pm.
What is the lowest common multiple?The smallest common positive number that is a multiple of two or more numbers.
Given that, a bell tolls every 10 minutes. another bell tolls every 15 minutes. both bells toll at 6:00pm.
To find the time for which both will ring together, we will find LCM of 10and 15
10 = 5x2
15 = 5x3
LCM = 5x2x3 = 30
The LCM of 15 and 10 is 30
So, both will ring after 30 minutes.
Right now the time is 6:00 pm, so after 30 minutes it will be 6:30 pm
Hence, Both the bells will ring after 30 minutes at 6:30 pm.
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Which point satisfies both f(x)=2^x and g(x)=3^x (0,1) (0,-1) (1,0) (-1,0)
Tablets are on sale for 15% off the original price (t), which can be expressed with the function p(t) = 0.85t. Local taxes are an additional 8% of the discounted price (p), which can be expressed with the function c(p) = 1.08p. Using this information, which of the following represents the final price of a tablet with the discount and taxes applied based on its original price? (2 points) c[p(t)] = 0.918t c(p) + p(t) = 1.93t c(p) ⋅ p(t) = 0.918pt t[c(p)] = 1.93p
The final price of the tablet is c(t) = 0.918.
What is sale price?
A sale price is the discounted price at which goods or services are being sold.
According to the given question
Tablets are on sale for 15% off the original price(t), which can be expressed with the function:
p(t) = 0.85t
The additional taxes of 8% of the discounted price(p), which can be expressed as a function:
c(p) = 1.08p
Therefore,
The final price of a tablet with the discount and taxes applied based on its original price is given by
c(p(t)) = 1.08 ×p(t)
c(p(t)) = 1.08×(0.85t)
c(p(t)) = 0.918t
⇒ c(t) =0.918t
Hence, the final price of the tablet is c(t) = 0.918.
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Write the sum using summation notation
729 + 1000 + 1331 + 1728 + ... + n^3
The party store has a special on greetings cards. It charges $14 for 4 greeting cards and $1.50 for each additional card. Write an equation for the total cost of greeting cards in terms of the number of cards. Define your variables. What is the total cost of 9 greeting cards?
C=14 +1.5(x-4)
c= total cost
x = number of cards
cost for 9 cards:
x=9
c=14+1.5(9-4)
14+1.5(5) = 21.50
total cost = $21.50
Given: The coordinates of iscosceles trapezoid JKLM are J(-b, c), K(b,c), L(a,0), and M(-a,0).
Prove: The diagonals of an isosceles trapezoid are congruent.
As part of the proof, find the length of KM
A) a2+b2+c2
B) (-a+b)2+c2
C) (a+b)2+c2
Answer with explanation:
It is given that, coordinates of Isosceles trapezoid J K L M are J(-b, c), K(b,c), L(a,0), and M(-a,0).
To Prove: The diagonals of an isosceles trapezoid are congruent.
Proof:
Distance formula , that is distance between two points in x y plane is given by
[tex]=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
Where, [tex](x_{1},y_{1}),(x_{2},y_{2})}[/tex] are coordinates of two points in the plane.
Length of Diagonal J L
[tex]=\sqrt{(a+b)^2+(0-c)^2}\\\\=\sqrt{(a+b)^2+c^2}[/tex]
Length of Diagonal K M
[tex]=\sqrt{(a+b)^2+(0-c)^2}\\\\=\sqrt{(a+b)^2+c^2}[/tex]
So, we can see that,
J L = KM [tex]=\sqrt{(a+b)^2+(c)^2}[/tex]
Hence,The diagonals of an isosceles trapezoid are congruent.
So ,
[tex]KM=\sqrt{(a+b)^2+c^2}[/tex]
Option C
bacterial colonies can triple in size every 4 days. if you start with 150 bacteria microorganisms, how large would the colony be after 16 days?
Find the product of (x − 2i)^2.
The product of (x-2i)²=x²-4+i4x
What is the process of multiplication of complex numbers?The product or multiplication of two complex numbers is also a complex number. The formula for multiplying complex numbers is: (a + ib) (c + id) = (ac - bd) + i(ad + bc).
Given here (x-2i)² =(x-2i) (x-2i)
=x²-4+i4x
Hence, the product is x²-4+i4x
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(6.3 × 1011) ÷ (7 × 105)
In the following triangle, find the values of the angles B and B', which are the best approximations to the solutions of this ambiguous case.
Answer:
Option B. B = 70.05° B' = 109.95°
Step-by-step explanation:
By the sine rule in a given triangle
sin 45°/16.5 = sinB/22
1/(1.414×16.5) = sinB/22
sinB = 22/(1.414×16.5) = 0.94
[tex]B = sin^{-1}(0.94)[/tex]
B = 70.05°
Now we know B' = 180 - Supplementary angle of B'
and B = B' ( opposite angles of equal sides are equal)
B' = 180 - B = 180 - 70.05 = 109.95°
Therefore option B is the answer.
how much would $500 invested at 6% interest compounded annually be worth after 4 years
How many sixteenth notes would be needed to have the same duration as 3 quarter notes? Represent this as a fraction
Julio just bought a $267,900 house. He had a 20 year mortgage with a fixed rate of 5.875%. Julio's monthly payments are $1558.09. What percent of the purchase price was Julio's down payment?
Answer: c
Step-by-step explanation:
Perform the operation(s) and write the answer in simplest form.
( 1/2 + 1/3) / 15/4
A.) 3 1/8
B.) 53/90
C.) 2/9
D.) 8/75
Use ABC to find the value of sin A. See picture below. Thanks!
Five students visiting the student health center for a free dental examination during national dental hygiene month were asked how many months had passed since their last visit to a dentist. their responses were as follows. 5 18 12 24 28 assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program. (give the answer to two decimal places.)
To find for the value of the confidence interval, let us first calculate for the values of x and s, the mean and standard deviation respectively.
x = (5 + 18 + 12 + 24 + 28) / 5
x = 17.4 months
s = sqrt{[(5 – 17.4)^2 + (18 – 17.4)^2 + (12 – 17.4)^2 + (24 – 17.4)^2 + (28 – 17.4)^2]/(5-1)}
s = 9.21
The formula for the confidence interval is given as:
Confidence Interval = x ± t s / sqrt(n)
Where t can be taken from standard distribution tables at 95% level at degrees of freedom = n – 1 = 4, t = 2.132. Therefore:
Confidence Interval = 17.4 ± 2.132 * 9.21 / sqrt(5)
Confidence Interval = 17.4 ± 8.78
Confidence Interval = 8.62 months, 26.18 months
Final answer:
To construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist, use the sample mean and sample standard deviation. Calculate the confidence interval using the formula and given data.
Explanation:
In order to construct a confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program, we can use the sample mean and sample standard deviation. The formula for constructing a confidence interval for the mean is:
Confidence Interval = sample mean ± (critical value) × (sample standard deviation / √sample size)
Using the given data, the sample mean is 17.4 months and the sample standard deviation is 9.18 months. With a confidence level of 95%, the critical value is 2.776. Plugging these values into the formula, we get:
Confidence Interval = 17.4 ± (2.776) × (9.18 / √5)
Calculating this, we find that the 95% confidence interval for the mean number of months elapsed since the last visit to a dentist is approximately 4.81 to 29.99 months.
Math help needed. Thank you
On monday eliza read her book. on tuesday she read three times as long as she read on monday. on wednesday she read 20 minutes less than tuesday. on thursday she read for 20 minutes which was half as long as she read on wednesday. how many minutes did eliza read over the 4-day period
type the slope-intercept equation of the line that passes through the points (0,2) and (2,0) y={?]x+{ }
What is the correct name for this circle?
∠EFG and ∠GFH are a linear pair, m∠EFG=3n+19, and m∠GFH=55+33 What are m∠EFG and m∠GFH?
Find all values of $x$ such that $6= \dfrac{35}{x} -\dfrac{49}{x^2}$. If you find more than one value, then list your solutions in increasing order, separated by commas.
Answer:
[tex]x=\frac{7}{3} , \frac{7}{2}[/tex]
Step-by-step explanation:
[tex]6= \frac{35}{x} - \frac{49}{x^2}[/tex]
Now we need to solve for x
To get 'x' alone we make the denominators same
LCD = x^2
WE multiply the whole equation by x^2
[tex]6x^2 = 35x - 49[/tex]
Now we the equation =0, move all the terms to left hand side
[tex]6x^2-35x + 49=0[/tex]
Now we apply quadratic formula to solve for x
a= 6, b= -35 , c= 49
[tex]x= \frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x= \frac{-(-35)+-\sqrt{(-35)^2-4(6)(49)}}{2*6}[/tex]
[tex]x= \frac{35+-\sqrt{49}}{12}[/tex]
[tex]x= \frac{35+-7}{12}[/tex]
Now frame two equations , one with + and another with -
[tex]x= \frac{35+7}{12}[/tex] [tex]x= \frac{35-7}{12}[/tex]
[tex]x= \frac{42}{12}[/tex] [tex]x= \frac{28}{12}[/tex]
[tex]x= \frac{7}{2}[/tex] [tex]x= \frac{7}{3}[/tex]
So value of x= {7/3, 7/2}
Which Graph correctly represents x+2y≤4?
Answer with Step-by-step explanation:
We are given an inequality:
x+2y≤4
We have to determine its correct graph
In graph A and graph C (0,4) lies in shaded region but (0,4) does not satisfy the inequality(since, 0+2×4=8 which is not less than or equal to 4)
Hence, A and C are not the graph of this inequality
In graph D (0,2) does not lie in shaded area but it satisfies the inequalityHence, D is also not the graph of this inequality
Hence, correct graph of x+2y≤4 is:
Graph B
Explain how you would use a number line to find the absolute value of –12.
What is the median of 36, 14, 21, 56, and 10?
Mr. calloway is an algebra teacher. every class period he draws a piece of paper out of a hat without looking to determine the number of homework problems he will assign. each different color of paper represents a different number of homework problems. the hat contains 2 blue, 6 red, 10 yellow, and 7 purple pieces of paper.what is the probability that mr. calloway draws a purple piece of paper during the first class period and a blue piece of paper during the second class period if he replaces all pieces of paper before each drawing?
The probability that Mr. Calloway will draw a purple piece of paper during the first class period and a blue piece during the second class period with replacement is [tex]\frac{14}{25}[/tex].
The question involves calculating the probability of drawing a purple piece of paper during the first class period and a blue piece of paper during the second class period with replacement. To find this probability, we first calculate the probability of each event separately since the events are independent. Mr. Calloway's hat contains a total of 25 pieces of paper (2 blue, 6 red, 10 yellow, and 7 purple).
The probability of drawing a purple piece of paper in the first class period is the number of purple papers divided by the total number of papers:
P(purple) = Number of purple pieces / Total pieces = [tex]\frac{7}{25}[/tex]
Since the pieces are replaced, the probabilities in the second class period are the same as in the first. Thus, the probability of drawing a blue piece of paper in the second class is:
P(blue) = Number of blue pieces / Total pieces = [tex]\frac{2}{25}[/tex]
Because these events are independent, we multiply the probabilities of each event happening:
Total probability = P(purple) times P(blue) = ([tex]\frac{7}{25}[/tex]) times ([tex]\frac{2}{25}[/tex]) = [tex]\frac{14}{25}[/tex].
Therefore, the probability that Mr. Calloway draws a purple piece of paper during the first class period and a blue piece of paper during the second class period with replacement is [tex]\frac{14}{25}[/tex].