Help me please!!!! i'll give brainliest!!

PLEASE HELP ME WITH THIS!!!

I am Lyosha [343]

Answer:

all are showing that y is a function of x

7 0

3 years ago

WHAT IS 9.1/3576, SHOWING STEPS

irga5000 [103]

$\frac { 9.1 }{ 3576 } \\ \\ =\frac { \frac { 91 }{ 10 } }{ 3576 } \\ \\ =\frac { 91 }{ 10 } \cdot \frac { 1 }{ 3576 } \\ \\ =\frac { 91 }{ 35760 }$

3 0

3 years ago

A university financial aid office polled a random sample of 670 male undergraduate students and 617 female undergraduate student

ahrayia [7]

Answer:

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Male undergraduates:

670, 388 were employed. So

$p_M = \frac{388}{670} = 0.5791$

$s_M = \sqrt{\frac{0.5791*0.4209}{670}} = 0.0191$

Female undergraduates:

Of 617, 323 were employed. So

$p_F = \frac{323}{617} = 0.5235$

$s_F = \sqrt{\frac{0.5235*0.4765}{617}} = 0.0201$

Distribution of the difference:

$p = p_M - p_F = 0.5791 - 0.5235 = 0.0556$

$s = sqrt{s_M^2+s_F^2} = \sqrt{0.0201^2 + 0.0191^2} = 0.0277$

Confidence interval:

The confidence interval is given by:

$p \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower bound of the interval is:

$p - zs = 0.0556 - 1.645*0.0277 = 0.01$

The upper bound of the interval is:

$p + zs = 0.0556 + 1.645*0.0277 = 0.1012$

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

5 0

3 years ago

Answer the questions below pleaseeee , i will give brainliest if they're right

Alex

Use formula A=bh+ L(s1+s2+s3)
#1- 974
#2- 737

3 0

2 years ago

Read 2 more answers

Lucas is walking around the edge of the playground shown. How far will lukas have to walk when he finishes one loop around the p

Tanya [424]

Answer:

See Explanation