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mixas84 [53]
3 years ago
11

Need help now!!! School starts soon!!!

Mathematics
1 answer:
Hatshy [7]3 years ago
6 0

Answer:

Since 21 hundredths is 21over one hundred,21 hundredths as a Fraction is 21/100. If you divide 21 by one hundred you get 21 hundredths as a decimal which is 0.21. To get21 hundredths as a Percent, you multiply the decimal with 100 to get the answer of 20percent.

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Find the value of 7v-10 given that -11v-7=4. Simplify your answer as much as possible.
densk [106]

First thing to do is solve the given equation for v

-11v-7 = 4

-11v = 4+7

-11v = 11

v = 11/(-11)

v = -1

Once we know this, we can use it to compute the following

7v-10 = 7*(-1) - 10 = -7 - 10 = -17

--------------------------------

Answer: -17

4 0
3 years ago
A college is currently accepting students that are both in-state and out-of-state. They plan to accept three times as many in-st
ziro4ka [17]

Answer:

0 < x ≤ 100 and 0 < y ≤ 300

Step-by-step explanation:

THIS IS THE COMPLETE QUESTION BELOW;

college is currently accepting students that are both in-state and out-of-state. They plan to accept three times as many in-state students as out-of-state, and they only have space to accept 100 out-of-state students. Let x = the number of out-of-state students and y = the number in-state students. Write the constraints to represent the incoming students at the college.

0 < x ≤ 100 and 0 < y ≤ 300

x > 0 and y > 0

0 < x ≤ 100 and y > 300

0 < x and y < 100

SOLUTION

they only have space to accept 100 out-of-state students,which means that the Maximum number of out-of-state students that can be accepted is 100

Then x= 100(Maximum number of out-of-state students that can be accepted)

They plan to accept three times as many in-state students as out-of-state which means that

Y = 3x(Maximum number of in-state students)

Then we can deduced that the numbee out-of-state students that can be accepted can lyes between the range of 0 and 100 which means from interval 0 to 100

Which can be written as 0 < x ≤ 100

But we need to know the interval for the Maximum number of in-state students(Y), to do that we need to multiply the equation above by 3 since Y = 3x

0 < x ≤ 100

3× 0 = 0

3× X = X

3× 100= 300

Then 0 < 3x≤ 300

But we know that Y = 3x then substitute into last equation

We have

0 < y ≤ 300

ThenBthe constraints to represent the incoming students at the college is

0 < x ≤ 100 and 0 < y ≤ 300

8 0
3 years ago
If the distribution is really (5.43,0.54)
defon

Answer:

0.7486 = 74.86% observations would be less than 5.79

Step-by-step explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:

\mu = 5.43, \sigma = 0.54

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{5.79 - 5.43}{0.54}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

7 0
3 years ago
5. h(t) = 3 • 49 for t= -3
Novosadov [1.4K]

Answer:

what do you need here there has to be something to do here your just giving me a variable

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Researchers measured the bone mineral desnsity of the spines of 94 women who had taken teh drug CEE. The mean was 1.016 g/cm2, a
Valentin [98]

Answer:

The 95% confidence interval for the mean is between 0.985g/cm² and 1.047 g/cm².

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96*\frac{0.155}{\sqrt{94}} = 0.0310

The lower end of the interval is the mean subtracted by M. So it is 1.016 - 0.0310 = 0.985 g/cm²

The upper end of the interval is the mean added to M. So it is 1.016 + 0.0310 = 1.047 g/cm²

The 95% confidence interval for the mean is between 0.985g/cm² and 1.047 g/cm².

6 0
3 years ago
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