HELP!!!!

Each person in ellens family drinks 8 1/3 ounces of water. If 5 people are in ellen's family. how many ounces of water do they d

Svetach [21]

Answer:

41 2/3 oz of water

Step-by-step explanation:

All you have to do is multiply 8 1/3 by 5.

First, lets convert 8 1/3 into an improper fraction (a fraction where the numerator is greater than the denominator):

8 1/3 = 25/3

now, we can multiply 25/3 by 5 and get 125/3, which simplifies to 41 2/3

5 0

3 years ago

ONLY ANSWER IF YOU KNOW THE ANSWER JUST FOR POINTS. Thank you! :)

Ksivusya [100]

Since triangle DEF = triangle JKL, m<D = m<J, m<E = m<K, m<F = m<L.
m<F = m<L = 90 degrees
m<K = m<E = 5(m<D)
but m<E + m<D = 90 degrees [right angled triangle]
5(m<D) + m<D = 90 degrees
6(m<D) = 90 degrees
m<D = 90 / 6 = 15 degrees.

5 0

3 years ago

Read 2 more answers

Solve the system of equations.<br> -10x+3y=5 <br> x=y-4

blsea [12.9K]

Answer:

x=y-4 -10x+3y=5

y=x+4 y=10/3x+5/3

x+4 =10/3x+5/3

x=1

y=1+4 y=10/3(1)+5/3

y=5 y=5

so

x=1

y=5

(1,5)

5 0

3 years ago

An SRS of 350 high school seniors gained an average of x¯=21 points in their second attempt at the SAT Mathematics exam. Assume

Lynna [10]

Answer:

a) $21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

b) $ME=2.58\frac{52}{\sqrt{350}}=7.17$

c) $ME=2.58\frac{52}{\sqrt{100}}=13.42$

d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

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

$\bar X=21$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=52$ represent the population standard deviation

n=350 represent the sample size

99% confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

Part b

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{350}}=7.17$

Part c

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{100}}=13.42$

Part d

As we can see when we reduce the sample size we increase the margin of error so the best option for this case is:

D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

6 0

2 years ago

Please i beg of you, help

IgorC [24]

Larry is a smart man, and he’s correct. (x+3)(4x-5)

8 0

3 years ago