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3 years ago

Find m∠1 if m∠1 = 6x and m∠CAB = 11x+9 NEED HELP ASAP PLEASE!

Mathematics

1 answer:

leva [86]3 years ago

3 0

Answer:

<1 = 54

Step-by-step explanation:

< CAB = <1 + <2

We have to assume that AP is a perpendicular bisector or we cannot solve the problem.

That would mean that <1 = <2

11x+9 = 6x+6x

11x+9 = 12x

Subtract 11x from each side

9 = 12x-11x

9 =x

We want <1

<1= 6x

<1 = 6*9

<1 = 54

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According to government data, 75% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women

blagie [28]

Answer:

We are given that According to government data, 75% of employed women have never been married.

So, Probability of success = 0.75

So, Probability of failure = 1-0.75 = 0.25

If 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

We will use binomial

Formula : $P(X=r) =^nC_r p^r q^{n-r}$

At x = 2

$P(X=r) =^{15}C_2 (0.75)^2 (0.25^{15-2}$

$P(X=2) =\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{13}$

$P(X=2) =8.8009 \times 10^{-7}$

b. That at most 2 of them have never been married?

At most two means at x = 0 ,1 , 2

So, $P(X=r) =^{15}C_0 (0.75)^0 (0.25^{15-0}+^{15}C_1 (0.75)^1 (0.25^{15-1}+^{15}C_2 (0.75)^2 (0.25^{15-2}$

$P(X=r) =(0.75)^0 (0.25^{15-0}+15 (0.75)^1 (0.25^{15-1}+\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{15-2})$

$P(X=r) =9.9439 \times 10^{-6}$

c. That at least 13 of them have been married?

P(x=13)+P(x=14)+P(x=15)

$={15}C_{13}(0.75)^{13} (0.25^{15-13})+{15}C_{14} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})$

$=\frac{15!}{13!(15-13)!}(0.75)^{13} (0.25^{15-13})+\frac{15!}{14!(15-14)!} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})$

$=0.2360$

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3 years ago

Which function has an inverse that is also a function?

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Answer: theirs is no question

Step-by-step explanation:

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2 years ago

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lubasha [3.4K]

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2 years ago

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R=7s/3+k/3

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3 years ago

The table shows the outputs, y, for different inputs, x: Input (x) 6 12 15 25 Output (y) 14 15 15 16 Part A: Do the data in this

cluponka [151]

Answer:

Part A: No it is not a function

Part B: The equation

Part C: x = 3

Step-by-step explanation:

Part A - The table:

Input (x) Output (y)

6 14

12 15

15 15

25 16

This is a function because it has no repeating inputs values which have alternate outputs. Each input is assigned exactly one output.

Part B - The relation y = 7x - 15 has the value y = 27 when x = 6. You can find it by substituting into the equation.

y = 7(6) -15

y = 42 - 15

y = 27

The value of the relation in the table when x = 6 is y = 14. The equation has the greater value.

Part C: To find when y = 6, set it equal to 6 and solve for x.

6 = 7x - 15

21 = 7x

3 = x

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2 years ago