Help 1+1 need help help help

What's the area and perimeter of a right triangle with sides 5, 12 and 13( 5 and 12 are the base and height)

uysha [10]

Perimeter is 30
Area is also 30.

Add the numbers to get the perimeter and then use 1/2bh to get the area. <span />

7 0

3 years ago

Read 2 more answers

What percentage of the whole school are 7th grade students? Round to the nearest tenth of a

11111nata11111 [884]

Answer:

How many kids are there in the whole school and how many are seventh graders? This question is not complete. Please complete it thank you.

Step-by-step explanation:

6 0

2 years ago

1 3/5 •1 2/5= Find the value of each expression in lowest terms. That dot between the numbers is times

IrinaK [193]

First multiply them out to improper fractions 1 3/5 changes into 8/5 and 1 2/5 changes into 7/5 multiply them out and you should get 2 6/25.

8 0

3 years ago

Which is the correct factoring and solutions?

aleksklad [387]

Step-by-step explanation:

your answer would be the second light blue one

(2h+1)(h+3) x=-1/2,-5

8 0

2 years ago

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amo

Aloiza [94]

Answer:

$a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]$

Step-by-step explanation:

a. -Given that:

$n_m=41\ , \ \sigma_m=33, \bar x_m=135.67\\\\n_f=37. \ \ ,\sigma_f=20, \ \ \bar x_f=68.64$

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

$\bar x_m-\bar x_f\\\\\therefore \bigtriangleup\bar x=135.67-68.64\\\\=67.03$

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator, $\bar x_m-\bar x_f$ is calculated by the following following:

$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

-And the margin of error, E at a 99% confidence can be calculated as:

$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

c. The estimator confidence interval is calculated using the following formula:

$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:

$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$

Hence, the 99% confidence interval is [51.2884,82.7716]

7 0

2 years ago