A basketball player scored x, y, and z points in 3 games. Express the average number of points scored by the player in terms of x, y, and z.

Answer: The value of x is 2 units.

Explanation:

Since we have given that

AB and CD are two chords that are intersected externally at point E.

As we know the relation regarding the above i.e.

[tex]EA\times EB=EC\times ED[/tex]

Here,

EB=x+1

EA=x+1+11 = (x+12)

EC=x+4

ED=x+4+1 = (x+5)

Now, we put the values in the relation given above :

[tex](x+1)(x+12)=(x+4)(x+5)\\\\x^2+13x+12=x^2+9x+20\\\\12+13x=20+9x\\\\13x-9x=20-12\\\\4x=8\\\\x=\frac{8}{4}=2\\\\x=2\ units[/tex]

Hence, the value of x is 2 units.

Answer:

x = 2 is the answer.

Step-by-step explanation:

When two chords AE and ED are drawn from an external point E to a circle then by theorem we know the relation

AE×BE = DE×CE

AE = x + 1 + 11 = x + 12

BE = x + 1

CE = x + 4

DE = x + 4 + 1 = x + 5

Now putting these values in the formula

(x + 12)(x + 1) = (x + 4)(x + 5)

x² + 12x + x + 12 = x² + 5x + 4x + 20

13x + 12 = 9x + 20

13x - 9x = 20 - 12

4x = 8

x = 2

x = 2 is the answer.

Answer: 95%

Step-by-step explanation:

A P E X

Option A. is the correct answer. Using the empirical rule and the provided narrow interval, the confidence level for the confidence interval of 46.6% to 47.4% is most likely 68%.

The question is asking to find the confidence level for a given confidence interval. In the scenario, a building manager finds that 47% of the time, people turn off the lights when they leave the room, with a sample standard deviation of 4%. To calculate the confidence level for the interval of 46.6% to 47.4%, we'd typically use a Z-score or t-score formula along with the sample size to determine which confidence level corresponds to the given interval. However, the provided question does not give us enough information to perform a calculation; thus, it is likely that the question expects us to understand the empirical rule (also known as the 68-95-99.7 rule).

Based on the empirical rule, the intervals corresponding to one, two, and three standard deviations from the mean (for a normal distribution) contain approximately 68%, 95%, and 99.7% of the data, respectively. Since the given confidence interval, 46.6% to 47.4%, is quite narrow and is likely to correspond to a plus or minus one standard deviation from the sample mean, the correct confidence level would be approximately 68% (Answer A). This assumption is due to the small range of the interval (0.4%) compared to the sample standard deviation (4%).

We are given the equation of the savings (s) as:

s = 50 + 5 w + 20 m

Part A.

Coefficient = 5 and 20

Variable = w and m

Constant = 50

Part B.

w = 10

m = 2

Therefore the total savings is:

s = 50 + 5 (10) + 20 (2)

s = $140

Part C.

The coefficient and the constant values are considered as fixed value because they do not change. Therefore only the variable value would change, however the expression 50 + 5 w + 20 m would still be the same. That is only the values of the variable change.

To avoid getting splashed from puddles on the street, Brian should stand in the middle of the park. This would distance him completely from the immediate spray zone of passing vehicles, like being further away from the shore reduces the impact of sprays from waves.

To avoid getting splashed by cars driving through the puddles, Brian should stand in the middle of the park. By doing so, he establishes the maximum possible distance between himself and any part of the street. Staying as far away from the streets reduces the risk of any water reaching him when cars pass by.

Think of it like Jill delivering her flyers and retracing her steps: the farthest spot from her house during her journey was at the midpoint of her travel route.

Also consider Peter's driving scenario: The farther he is from other cars, the less likely he is to get affected by their maneuvers. In a similar way, Brian, being in the middle of the park would keep him away from the splash zone of passing cars.

Lastly, akin to Julian and his father on their surfboards, being away from the impact point of the waves (which is near the shore) they get least affected by the spray. This analogy can be compared to Brian's situation with the puddles.

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The total number of hours Vanessa work in a week is:

total work hours = (8 hours / day) * 5 day / week

total work hours = 40 hours per week

The total number of hours she did not work:

total hours did not work = (90 minutes)(1 hour / 60 minutes) + (45 minutes)(1 hour / 60 minutes)  + 2.5 hours

total hours did not work = 4.75 hours

The amount of hours Vanessa work last week can be calculated using the formula:

work hours = total work hours - total hours did not work

work hours = 40 hours – 4.75 hours

work hours = 35.25 hours

the last digit is the ones place, if it is 5 or higher you would round up

 under 5 you round down

 so to the nearest 10 the number would be 67910

You are given the mass of CaH₂ which is 14 grams and H₂O which is 28 grams. You are required to find the maximum volume of H₂ gas at STP. The balanced chemical reaction is CaH₂ + 2H₂O → Ca(OH)₂ + H₂. Note that for every one mole of CaH₂, 2 moles of H₂O is needed to completely react and produce Ca(OH)₂ + H₂. So the CaH₂ and H₂O are at equal proportions. At STP, temperature is at 0°C(273K) and pressure at 1 atm. The molar mass of CaH₂ is 42 grams per mole.

14g CaH₂(1mol CaH₂/42 grams CaH₂)(1mol H₂/1mol CaH₂) = 0.333 moles H2

The ideal gas equation is PV = nRT, with R(gas constant) = 0.08206 L-atm/mol-K. get the equation for volume, we have

V = nRT/P

V = (0.333mol H₂)(0.08206 L-atm/mol-K)(273K)/1atm

V = 7.47L

The answer is a positive integer.

Option b is the correct answer. Exponents should be multiplied in the expression (1/3)^4*(1/3)^7, resulting in (1/3)^11.

The expression where the exponents should be multiplied is Option b. (1/3)^4*(1/3)^7. This is because when you multiply two expressions that have the same base, you add their exponents. In this case, you have the base (1/3) raised to the 4th power and then to the 7th power, so you add the exponents 4 + 7, which equals 11. Therefore, (1/3)^4*(1/3)^7 simplifies to (1/3)^11.

Another case where you multiply exponents would be when raising a power to another power, like in (5^3)^4. Here the exponents 3 and 4 are multiplied, resulting in (5^12).

5, 12, 13 are the sides representing lengths of sides of a right triangle.

The theorem attributed to Pythagoras that the square on the hypotenuse of a right-angled triangle is equal in area to the sum of the squares on the other two sides.

Given that, lengths of sides of triangles.

If a triangle is a right triangle, then it should follow Pythagoras theorem,

Using Pythagoras theorem,

13² = 169

12² + 5² = 144 + 25 = 169

Therefore, 13² = 12² + 5² is following Pythagoras theorem.

Hence, 5, 12, 13 are the sides representing lengths of sides of a right triangle

For more references on Pythagoras theorem, click;

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Final answer:

To find the nth term equation of an arithmetic sequence, we first determined the common difference to be 92 by given terms, and then found the first term by using the formula for arithmetic sequences. The nth term equation for the sequence is: an = -1489 + (n-1)(92).

Explanation:

To find the nth term equation of the arithmetic sequence, we need to determine the common difference and the first term of the sequence. Since we know the 18th term, a18 = 97, and the 20th term, a20 = 281, we can find the common difference, d, by subtracting the two given terms and dividing by the number of terms between them:

d = (281 - 97) / (20 - 18) = 184 / 2 = 92.

Once we have the common difference, we can use the following formula to find the first term (a1):

an = a1 + (n-1)d.

Let's substitute n = 18 and d = 92 into the formula:

97 = a1 + (18-1)(92)

a1 = 97 - (17)(92) = -1489.

Now we can write the equation for the nth term of the arithmetic sequence:

an = -1489 + (n-1)(92).

Answer:

[tex]x=-1+\sqrt{23}\; ,\;-1-\sqrt{23}[/tex]

Step-by-step explanation:

Roots of quadratic equations [tex]ax^{2}+bx+c=0[/tex] calculated as:

[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]

First, compare [tex]ax^{2}+bx+c=0[/tex] with [tex]x^{2}+2x-22=0[/tex]

here a = 1 , b = 2 and c = -22

Now, plug the value of a,b and c in [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]

[tex]x=\frac{-2\pm \sqrt{2^{2}-4(1)(-22)}}{2(1)}[/tex]

[tex]x=\frac{-2\pm \sqrt{4+88}}{2}[/tex]

[tex]x=\frac{-2\pm \sqrt{92}}{2}[/tex]

[tex]x=\frac{-2\pm 2\sqrt{23}}{2}[/tex]

[tex]x=-1\pm \sqrt{23}[/tex]

Hence, [tex]x=-1+\sqrt{23}\; ,\;-1-\sqrt{23}[/tex]