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morpeh [17]
4 years ago
7

Help Me?????????????

Mathematics
1 answer:
Rzqust [24]4 years ago
7 0
What you have to do is multiply 16 times 6 = 96
96x96= 9216
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What is the value of x in trapezoid ABCD? x=15 x=20 x=45 x=60
zvonat [6]

Answer:

A. <u><em>X = 15</em></u> is the correct answer.

Step-by-step explanation:

It's the only one that really makes sense.

Hope this helped :)

6 0
3 years ago
Read 2 more answers
5x^{4}-7x^{3}-5x^{2}+5x+1-0
den301095 [7]

Answer:

5x^4 -7x^3-5x^2+5x+1

Step-by-step explanation:

Hey!

You would just add all the numbers up, and by making sure that you combine all like terms. In this case it is just 1-0 = 1 and so the solution would be the same as your question but with a 1 and you would get rid of the -0.

Hope this helps!

7 0
3 years ago
What is the area of a regular nonagon (9 sides) with side length of 15 cm and apothem of 20.6 cm? Round the answer to the neares
Triss [41]
We know that

<span>The nine radii of a regular Nonagon divides into 9 congruent isosceles triangles
</span>therefore
[the area of <span>a regular nonagon]=9*[area of isosceles triangle]

</span>[area of isosceles triangle]=b*h/2------> 15*20.6/2----> 154.5 cm²
so
[the area of a regular nonagon]=9*[154.5]------> 1390.5 cm²

the answer is
1390.5 cm²

5 0
3 years ago
A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec
Ket [755]

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

(a)

The null hypothesis is:

<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.

Assume that the significance level of the test is, <em>α</em> = 0.05.

Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

For 5% level of significance the two tailed critical value of <em>z</em> is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct the 95% confidence interval as follows:

CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)

As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

3 0
3 years ago
Read 2 more answers
Help me on number problem 4 please.
Tresset [83]
The answer is 460 because you have to subtract
3 0
3 years ago
Read 2 more answers
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