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3 years ago

Which numbers repeat in the decimal form of 5/11

Mathematics

1 answer:

sladkih [1.3K]3 years ago

6 0

Answer:

0.45 repeating

Step-by-step explanation:

you divide 5 and 11 then get 0.45 repeating

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Donna bought 3 pairs of pants for $28.50 each. She also bought a necklace for $12.25. She received a discount of 35% off her ent

oee [108]

Answer:

<em>Donna will have to pay $67.985125 at the register, including the 7% sales tax.</em>

Step-by-step explanation:

Donna bought 3 pairs of pants for $28.50 each.

So, the cost for 3 pair of pants $=(28.50\times 3)dollars= 85.50$ dollars.

She also bought a necklace for $12.25

Thus, the cost for her entire purchase will be: $(85.50+12.25) dollars = 97.75$ dollars.

She received a discount of 35% off her entire purchase. So, the amount of discount $=(97.75\times 0.35)= 34.2125$ dollars.

Thus, the total selling price will be: $(97.75-34.2125)= 63.5375$ dollars.

Percentage of sales tax is 7%. So, the amount of sales tax $=(63.5375\times 0.07)= 4.447625$ dollars.

Thus, the total amount Donna have to pay at the register $=(63.5375+4.447625)=67.985125$ dollars.

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4 years ago

What is the radius of a circle with an area of 50.24 square inches?

Vladimir [108]

Answer:

<h3>Area of a circle= πr²</h3><h2>50.24 = 3.14 *r²</h2><h2>r²= 50.24/3.14</h2><h2>r² =16</h2><h2>r= √16</h2><h2>r = 4</h2>

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3 years ago

For a function, the set of all possible values that can be used for the independent variable is called the

Bess [88]

The independent variable (x) is the domain

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3 years ago

Question 95 pts

garri49 [273]

The correct answer is “hasn’t” probably maybe

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3 years ago

Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra

Kobotan [32]

<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = sample mean

n= sample size

s= sample standard deviation

$t_{\alpha/2}$ = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , $\overline{x}=22.4$

$\alpha=1-0.99=0.01$

A) Critical t value for $\alpha/2=0.005$ and degree of 18 freedom

$t_{\alpha/2}$ = 2.8784

B) Required confidence interval:

$22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)$

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

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3 years ago