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3 years ago

Mary has walked 8 km, which is 55% of the total distance she will walk. How much farther

Mathematics

1 answer:

Viefleur [7K]3 years ago

3 0

To find the total distance, Divide distance already walked by percentage:

8 / 0.55 = 14.55 km

She already walked 8 so now subtract:

14.55 - 8 = 6.55 km left to walk

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Add the numerator and denominator seperatly, then find the quotient <br> -3( - 8) / 5 + 7

kipiarov [429]

24/ 12=2 because -3 times -8 is positive 24 divided by 5+7 which is 12 and 12 goes into 24 twice

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3 years ago

A researcher is comparing the effectiveness of three devices designed to help people who snore. There are 60 people who snore pa

aleksandr82 [10.1K]

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2 years ago

What is the length of the hypotenuse? If necessary, round to the nearest tenth.

Brut [27]

Answer:

c ≈ 5.5 cm

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

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c = $\sqrt{29.77}$ ≈ 5.5 cm ( to the nearest tenth )

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3 years ago

Read 2 more answers

the linear combination method is applied to a system of equations as shown.4(.25x .5y = 3.75) → x 2y = 15 (4x – 8y = 12) → x – 2

gtnhenbr [62]

If we use the last equation:
2 x = 18
x = 18 : 2
x = 9
Then we will put it in another equation:
x - 2 y = 3
9 - 2 y = 3
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4 years ago

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there

yan [13]

Answer:

Explained below.

Step-by-step explanation:

According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normally.

Then, the mean of the sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The expected value of the sample mean of their weights is same as the population mean, <em>μ</em> = 1515 lbs.

The standard deviation of the sampling distribution of the sample mean weight is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{25}{\sqrt{16}}=6.25$

The average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2500 lbs. is:

$\text{Average Weight}=\frac{2500}{16}=156.25$

Compute the probability that a random sample of 16 persons on the elevator will exceed the weight limit as follows:

$P(\bar X > 156.25)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{156.25-151}{6.25})\\\\=P(Z>0.84)\\\\=1-P(Z$

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3 years ago