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Mekhanik [1.2K]
3 years ago
14

Help me please,Thanks

Mathematics
1 answer:
Alborosie3 years ago
4 0

the formula for volume of a prism is base x height
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1064 x

19

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20216 legos

Sharon has 20216 legos

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2 years ago
What exclusions are placed on the variable x for the fraction 4x^2-1/4x+2
aalyn [17]
Below are the choices:

<span>-1/2
0
 -1
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The  exclusions are placed on the variable x for the fraction 4x^2-1/4x+2 is -1/2. Below is the solution, 

<span>4x+2 = 0 so 4x=-2 so x=-2/4 wich is -1/2</span>
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3 years ago
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PLS HELP ME I SUCK AT MATH
Rainbow [258]

Divide the sides of the larger triangle by the corresponding side of the smaller one:

18/7.2 = 2.5

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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
2 years ago
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Solve for a<br><br> =50=15a<br> Simplify your answer as much as possible.
Ivan
3.3333333333..... is the answer
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3 years ago
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