Write 2.1111111 as a mixed number

The value of a is -1

What do perpendicular lines have a slope of?

Perpendicular lines have slopes that are negative reciprocals of each other. The given line's slope is 5, which means that the slope of the opposite line must be its negative reciprocal.

Conclusion: The slopes of two perpendicular lines are negative reciprocals of each other. This means that if a line is perpendicular to a line that is the slope of the line is -1/m.

let the slope of line AB is m1 and the slope of line BC be m2

m1= 3+3/a-2 = 6/a-2

m2=1+2/10-2 = 1/2

m1 * m2 = -1

6/a-2 * 1/2 = -1

3 = -a+2

a = -1

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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?

Answer:

The answer is all integers from 1 to 10, inclusive

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Answer:

2.828427

Step-by-step explanation:

I looked it up

To find the volume of the solid described, we can use the formula for the volume of a cylinder and integrate the cross-sectional area. The cross-sectional area is the base of the solid, and in this case, it is given by (25 - 50x^2 + 25x^4).

The solid described in the question has a base in the shape of the region enclosed by the parabola y = 5 - 5x^2 and the x-axis. The cross-sections perpendicular to the x-axis are isosceles triangles with a height equal to the base.

To find the volume of the solid, we can use the formula for the volume of a cylinder. The cross-sectional area of each triangle is the base times the height, and the height of the triangle is also the base. So, the cross-sectional area is A = (5 - 5x^2)(5 - 5x^2) = (25 - 50x^2 + 25x^4).

The volume of the solid is then given by integrating the cross-sectional area from the x-values that define the base of the solid. So, V = ∫(25 - 50x^2 + 25x^4)dx.

Divide 5821 by 71 

Round the number to the nearest hundred

5821 ÷ 71 → 5800 ÷ 70 

Calculate mentally. 

58 ÷ 7=8.2

Estimated quotient = 8

combine all the like terms

 8x^2 + x^2 = 9x^2

-9y^2 - 3y^2 = -12y^2

-4x-7x = -11x

 so you have 9x^2 -12y^2-11x

 Answer id the last one

The expression as the product of two factors is (1 + 7)(2 + 1).

For the given question, the expression is;

=  2x1 + (1+7) + (7x2)

First, search for numbers that share a factor.

Consider the factors (2 x 1) and (7 x 2); this same common factor is two, so we factor.

= (2 x 1) + (7 x 2) + (1 +7)

Taking 2 common from first two numbers.

= 2 (1 + 7 ) + (1 + 7)

Now, take (1+7) common from both.

= (1 + 7)(2 + 1)

Thus, the expression as the product of two factors is (1 + 7)(2 + 1).

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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

Hiya :)

d. exponential growth

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Hope this helped!!

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(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.