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Yuki888 [10]
2 years ago
11

An equilateral triangle has a semiperimeter of 6 meters.

Mathematics
2 answers:
Ierofanga [76]2 years ago
8 0

Answer:

B

Step-by-step explanation:

Took the test

guapka [62]2 years ago
7 0

Answer:

7

Step-by-step explanation:

An equilateral triangle has a semiperimeter of 6 meters.

Heron’s formula: Area = StartRoot s (s minus a) (s minus b) (s minus c) EndRoot

An equilateral triangle has a semiperimeter of 6 meters. What is the area of the triangle? Round to the nearest square meter.

2 square meters

7 square meters

20 square meters

78 square meters

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The box plot compares Marta’s and Ani’s diving scores in the first several meets of the season.
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Answer:

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If you plot the data (2, 4, 6, 8, and 12) it looks something like the poor box-and-whisker plot below

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Description A describes that exact box-and-whisker plot.

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the population of Ketcham high school has been decreasing by 5 % per year.if it's population is currently 2600 which of the foll
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A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec
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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

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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.

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