How many diameters can a circle have?

297 /3 = 99

99-1 = 98

99 +1 = 100

98 + 99 + 100 = 297

 the numbers are 98, 99 , 100

Answer: 7 + 7 = 14

8 + 8 = 16

Step-by-step explanation: doubles facts are simply additions where a number is added to it self. The strategy sums up two consecutive numbers when they are next to each other to give their result as given by the question above (8 + 7). We simply add the smaller number together then add one (double-plus-one) OR add the larger number together then subtract one (double-minus-one)

All doubles that can be used in solving 8 + 7 are:

A) 8 + 7 = 7 + (7 + 1) = (7 + 7) + 1 = 14 + 1 = 15 [double-plus-one]

B) 8 + 7 = (8 + 8) - 1 = 16 - 1 = 15 [double-minus-one]

The doubles fact makes use of the associative property of addition —changing the grouping of addends does not change the sum.

The required function notations for the total law firm charges is expressed as f(t) = 100t + 300

Given the following

Law firm charges = $100 per hour

The amount of charge for "t" hours will be 100t hours

Also, the original fee = $300

In other to get the total charge using function notation;

f(t) = Law firm charges for "t" hours + Original fee

f(t) = 100t + 300

The required function notations for the total law firm charges is expressed as f(t) = 100t + 300

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Answer:

2.828427

Step-by-step explanation:

I looked it up

Answer:

The answer is all integers from 1 to 10, inclusive

Bye~~

Answer:

a) 5.2 *10^7 km

Step-by-step explanation:

If we could describe our Solar System, in order of appearance nearer the Sun, it would be like this:

Sun --- Mercury --- Venus --- Earth

Sun --------------------- Venus

1.1 * 10^8 km

Sun -------Mercury

5.8 * 10^7 Km

To find out how many more kilometers is the distance of Venus from the Sun than the distance of Mercury from the Sun, all we have to do is simply subtract the distance Sun ---Venus minus Sun ---Mercury

So,

1.1 * 10^8 - 5.8 * 10^7 =

Adjusting the first distance to the same power

110*10^7- 5.8*10^7 =

Subtracting the factors

5.2 * 10^7

(a) The probability is [tex]\[\boxed{0.464}\][/tex]. (b) The probability is [tex]\[\boxed{0.763}\][/tex]. (c) The probability is [tex]\[\boxed{0.585}\][/tex]. (d) The probability is [tex]\[\boxed{0.921}\][/tex]. (e) No, because 205 females are frequently involved in charity work. The option (A) is correct.

To address the given questions based on the provided table, let's go through each question step-by-step:

(a) Find the probability that the person is frequently or occasionally involved in charity work.

First, we need the total number of people who are frequently or occasionally involved in charity work. This is the sum of people in the "Frequently" and "Occasionally" columns.

[tex]\[\text{Total frequently or occasionally involved} = 432 + 904 = 1336\][/tex]

Now, we divide this by the total number of people surveyed:

[tex]\[P(\text{frequently or occasionally involved}) = \frac{1336}{2881} \approx 0.464\][/tex]

So, the probability is [tex]\[\boxed{0.464}\][/tex].

(b) Find the probability that the person is female or not involved in charity work at all.

To solve this, we need to find the number of females and those not involved in charity work at all.

[tex]\[\text{Total females} = 1402\][/tex]

[tex]\[\text{Total not involved at all} = 1545\][/tex]

We need to subtract the overlap (females not involved in charity work) to avoid double-counting. From the table, the number of females not involved at all is 747.

[tex]P(\text{female or not involved at all}) = \frac{\text{Total females} + \text{Total not involved at all} - \text{Females not involved}}{\text{Total}}[/tex]

[tex]= \frac{1402 + 1545 - 747}{2881} = \frac{2200}{2881} \approx 0.763[/tex]

So, the probability is [tex]\[\boxed{0.763}\][/tex].

(c) Find the probability that the person is male or frequently involved in charity work.

[tex]\[\text{Total males} = 1479\][/tex]

[tex]\[\text{Total frequently involved} = 432\][/tex]

We need to subtract the overlap (males frequently involved) to avoid double-counting. From the table, the number of males frequently involved is 227.

[tex]P(\text{male or frequently involved}) = \frac{\text{Total males} + \text{Total frequently involved} - \text{Males frequently involved}}{\text{Total}}[/tex]

[tex]= \frac{1479 + 432 - 227}{2881} = \frac{1684}{2881} \approx 0.585[/tex]

So, the probability is [tex]\[\boxed{0.585}\][/tex].

(d) Find the probability that the person is female or not frequently involved in charity work.

[tex]\[\text{Total females} = 1402\][/tex]

[tex]\[\text{Total not frequently involved} = 2881 - 432 = 2449\][/tex]

We need to subtract the overlap (females not frequently involved) to avoid double-counting. From the table, the number of females not frequently involved is 1197 (450 + 747).

[tex]P(\text{female or not frequently involved})=\frac{1402 + 2449 - 1197}{2881} = \frac{2654}{2881} \approx 0.921[/tex]

So, the probability is [tex]\[\boxed{0.921}\][/tex].

(e) Are the events "being female" and "being frequently involved in charity work" mutually exclusive?

Two events are mutually exclusive if they cannot occur at the same time.

From the table, 205 females are frequently involved in charity work.

Since there are females who are frequently involved in charity work, the events "being female" and "being frequently involved in charity work" are not mutually exclusive.

So, the answer is A. No, because 205 females are frequently involved in charity work.

The complete question is:

The table below shows the results of a survey that asked 2881 people whether they are involved in any type of charity work. A per selected at random from the sample. Complete parts (a) through (e).

(a) Find the probability that the person is frequently or occasionally involved in charity work.

P(being frequently involved or being occasionally involved) - (Round to the nearest thousandth as needed.)

(b) Find the probability that the person is female or not involved in charity work at all.

P(being female or not being involved) (Round to the nearest thousandth as needed.)

(c) Find the probability that the person is male or frequently involved in charity work.

P(being male or being frequently involved) (Round to the nearest thousandth as needed.)

P(being male or being frequently involved) - (Round to the nearest thousandth as needed.)

(d) Find the probability that the person is female or not frequently involved in charity work.

P(being female or not being frequently involved) = (Round to the nearest thousandth as needed.)

(e) Are the events "being female" and "being frequently involved in charity work" mutually exclusive? Explain.

A. No, because 205 females are frequently involved in charity work.

B. Yes, because no females are frequently involved in charity work.

C. Yes, because 205 females are frequently involved in charity work.

D. No, because no females are frequently involved in charity work.

The probability of having a baby boy on the fourth child, given that the first three children were all boys, is 0.5. This result is not the same as the probability of getting all boys among four children, which is 0.0625. The conditional probability accounts for the information about the first three births.

To solve this probability problem, let's break it down step by step.

Probability of Having a Boy on the Fourth Child:

Assuming boys and girls are equally likely, the probability of having a boy or a girl is 1/2 or 0.5. When considering each child's gender independently, the probability of having a boy on the fourth child is 0.5, regardless of the genders of the previous children.

However, the question specifies that the first three children were all boys. This information is crucial for the conditional probability calculation.

Conditional Probability:

The probability of having a boy on the fourth child given that the first three children were all boys is denoted as [tex]\( P(B_4 | B_1, B_2, B_3) \)[/tex].

Since the events are assumed to be independent (the gender of one child does not affect the gender of another), the conditional probability is the same as the probability of having a boy on any single birth: 0.5.

Comparison with Getting All Boys:

The probability of getting all boys among four children [tex](\( P(B_1 \cap B_2 \cap B_3 \cap B_4) \))[/tex] is the product of the probabilities of having a boy for each birth.

[tex]\[ P(B_1 \cap B_2 \cap B_3 \cap B_4) = P(B_1) \times P(B_2) \times P(B_3) \times P(B_4) \][/tex]

Given that [tex]\( P(B_4) = 0.5 \)[/tex] and the previous births are all boys, [tex]\( P(B_1 \cap B_2 \cap B_3 \cap B_4) = (0.5)^4 = 0.0625 \)[/tex].

The question probable may be:

Find the probability of a couple having a baby boy when their fourth child is born, given that the first three children were all boys. Assume boys and girls are equally likely. Is the result the same as the probability of getting all boys among four children?

Option: A is the correct answer.

        A girl is 9; mom is 27

A girl is now one-third as old as her mother.

i.e. if x is the present age of girl.

and y is the present age of her mother.

Then,

[tex]x=\dfrac{1}{3}y[/tex]

i.e.

[tex]y=3x-----------(1)[/tex]

In three years, she will be two-fifths as old as her mother will be.

This means after three years.

The age of girl will be: x+3

and the age of her mother will be: y+3

This means that:

[tex](x+3)=\dfrac{2}{5}\times (y+3)[/tex]

[tex]5(x+3)=2(y+3)\\\\i.e.\\\\5x+15=2y+6[/tex]

i.e.

[tex]5x+15=2\times 3x+6[/tex]

( since on using equation (1) )

i.e.

[tex]5x+15=6x+6\\\\i.e.\\\\6x-5x=15-6\\\\i.e.\\\\x=9[/tex]

and the value of y from equation (1) is:

[tex]y=27[/tex]

The statement which is equivalent to line segment CD is congruent to line segment XY is CD overbar is congruent to XY overbar.

A line is made up of an infinite no. of points it can extend in both directions indefinitely.

We know a line has two subsets they are a ray and a line segment.

A ray is a type of line that has one initial point and the other end can extend indefinitely and a line segment is a type of line which has two endpoints.

Given a line, segment CD is congruent to line segment XY.

∴ [tex]\overline{CD}[/tex] ≅ [tex]\overline{XY}[/tex].

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