State the set of multiples of 14 between 7 and 84 inclusive

What is the approximate area of sector BAC given that 0 ~ 4.71 radians?

IRISSAK [1]

Answer:

1962.5 m^2

Step-by-step explanation:

area of the entire circle = pi (50^2)

subtract the large area

pi 50^2 - 4.71 /2pi pi 50^2 =

7 0

3 years ago

A magazine reports that women trust recommendations from a particular social networking site more than recommendations from any

Misha Larkins [42]

Answer:

a) $p_A$ represent the real population proportion for women

$\hat p_A =\frac{123}{150}=0.82$ represent the estimated proportion for women

$n_A=150$ is the sample size selected for women

b) $p_B$ represent the real population proportion for male

$\hat p_B =\frac{102}{170}=0.6$ represent the estimated proportion for male

$n_B=170$ is the sample size selected for male

c) $(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124$

$(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316$

And the 95% confidence interval would be given (0.124;0.316).

We are confident at 95% that the difference between the two proportions is between $0.124 \leq p_A -p_B \leq 0.316$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

$p_A$ represent the real population proportion for women

$\hat p_A =\frac{123}{150}=0.82$ represent the estimated proportion for women

$n_A=150$ is the sample size selected for women

Part b

$p_B$ represent the real population proportion for male

$\hat p_B =\frac{102}{170}=0.6$ represent the estimated proportion for male

$n_B=170$ is the sample size elected for male

$z$ represent the critical value for the margin of error

Part c

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124$

$(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316$

And the 95% confidence interval would be given (0.124;0.316).

We are confident at 95% that the difference between the two proportions is between $0.124 \leq p_A -p_B \leq 0.316$

8 0

3 years ago

Identify the horizontal asymptote of f(x) = 7x+1/2x-9

horrorfan [7]

Answer:

D

Step-by-step explanation:

3 0

3 years ago

4. What is the value of the expression 72 + (33 + 12) 23 a. 3 b. 8 C. 9 d. 11

jeyben [28]

Answer:

1107

Step-by-step explanation:

72+ (33+12)23

= 72 + (45)23

=72+ 1035

=1107

6 0

3 years ago

2 At Bea's Pet Shop, the number of dogs, d, is initially five less than twice the number of cats, c. If she decides to add three

KatRina [158]

Let D be dogs and C be cats

dogs, d, is initially five less than twice the number of cats, c

D + 5 = 2C

If she decides to add three more of each, the ratio of cats to dogs will be

D + 8 = 2C + 3

Could Bea's Pet Shop initially have 15 cats and 20 dogs?

Simply plug in the numbers

20 + 5 = 2(15)

This is clearly not true: 25 does not equal 30

6 0

3 years ago

State the set of multiples of 14 between 7 and 84 inclusive​

State the set of multiples of 14 between 7 and 84 inclusive