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4 years ago

10

Kendra has 3/4 of a bag of hair ribbons. She wants to share so that each of her friends gets 3/8 of the box. How many friends ca

n Kendra give hair ribbons to?

1 answer:

aalyn [17]4 years ago

7 0

3/4 = 6/8 and 3/8 is half of 6/8. If she wanted to give each friend 3/8 of the bag, she could give a portion to only 2 friends.

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Andrew is buying pencils, and he wants to buy fewer than 25. He plans to give 75% of the pencils to his students and save three

sweet-ann [11.9K]

Answer:

50%

Step-by-step explanation:

the will have 75minis 25

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3 years ago

Read 2 more answers

ANSWER PLEASE *50 POINTS*

zvonat [6]

If a football player runs 5/6 of a mile every 3/4th of a hour, that means every 1/4th a hour he runs 1.5/6th of a mile.

That means every 15 minutes the football player runs .2 of a mile. That means that every hour he runs :

0.8 of a mile!

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3 years ago

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

ziro4ka [17]

Answer:

a) $\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=9-1=8$

The Confidence interval is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and the critical values are:

$\chi^2_{\alpha/2}=20.09$

$\chi^2_{1- \alpha/2}=1.65$

And replacing into the formula for the interval we got:

$\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

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4 years ago

Which is a better buy, 4 lunches for $57 or 5 lunches for $71?

worty [1.4K]

Answer: 5 lunches for $71 is better
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71/5=14.2

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3 years ago

Describe the end behavior of the function f(x)=-2x^4- x^3 +3.

Hoochie [10]

Answer:

B

Step-by-step explanation:

For a polynomial , the behaviour as x gets large is determined by the term of greatest degree,

f(x) = - 2 $x^{4}$ - x³ + 3

Is of even degree with a negative leading coefficient , then

f(x) → - ∞ as x → + ∞ and f(x) → - ∞ as x → - ∞

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2 years ago