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Oksanka [162]
2 years ago
5

Which expression is equivalent to

n\frac{7\pi }{6}" alt="sin\frac{7\pi }{6}" align="absmiddle" class="latex-formula">?

Mathematics
1 answer:
Ulleksa [173]2 years ago
7 0

Check the picture below.

notice, the pairs in the Unit Circle are the (cosine , sine) pair, which are equivalent to (x , y) values in a cartesian plane.

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I need some help please
9966 [12]
Dilation:

The dilation factor is the length of image/preimage = 1.5

Thus
Q'R'/QR=1.5
Q'R'=1.5QR = 1.5 (5 cm) = 7.5 cm

(Edited)

8 0
3 years ago
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Can anyone help with this problem?
dangina [55]
The answer is 48, because you do 24•2
4 0
3 years ago
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Write the numerical values of the following expression.<br> -3|x|+2x-1 if x=-5
Mademuasel [1]

Answer:

-26

Step-by-step explanation:

-3|-5| + 2(-5) - 1

-3(5) - 10 - 1

-26

6 0
4 years ago
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The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6.
Wittaler [7]

Answer:

a) There is a 10.75% probability of observing less than 60 hits in an hour.

b) The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) There is a 24% probability of observing between 80 and 90 hits an hour

Step-by-step explanation:

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}

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem, we have that

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so \mu = 75, \sigma = 8.6.

a) What’s the probability of observing less than 60 hits in an hour? Use the normal approximation

This is the pvalue of Z when X = 60. So

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

Z = \frac{60 - 75}{8.6}

Z = -1.74

Z = -1.74 has a pvalue of 0.1075. This means that there is a 10.75% probability of observing less than 60 hits in an hour.

b) What’s the 99th percentile of the distribution of the number of hits?

What is the value of X when Z has a pvalue of 0.99.

Z = 2.35 has a pvalue of 0.99

So

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

2.35 = \frac{X - 75}{8.6}

X - 75 = 20.21

X = 95.21

The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) What’s the probability of observing between 80 and 90 hits an hour?

This is the pvalue of the zscore of X = 90 subtracted by the pvalue of the zscore of X = 80.

For X = 90

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

Z = \frac{90 - 75}{8.6}

Z = 1.74

Z = 1.74 has a pvalue of 0.95907

For X = 80

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

Z = \frac{80 - 75}{8.6}

Z = 0.58

Z = 0.58 has a pvalue of 0.71904

So

There is a 0.95907 - 0.71904 = 0.24003 = 24% probability of observing between 80 and 90 hits an hour

6 0
3 years ago
A company produces packages of granola and has changed its recipe. It claims that the proportion of almonds in how is the approp
vova2212 [387]

Answer:

The correct answer is

t = \frac{0.49-0.29}{\sqrt{\frac{(0.42)^2}{100} +}\frac{(0.29)^2}{100}  }

Step-by-step explanation:

To solve the question we note that the question involves a test statistic calculation given by

t = \frac{x_1-x_2}{\sqrt{\frac{s_1^2}{n_1} +}\frac{s_2^2}{n_2}  } Where

x₁ = Mean of sample 1

x₂ = Mean of sample 2

n₁ = Sample size of sample 1

n₂ = Sample size of sample 2

s₁ = Variance of sample 1

s₂ = Variance of sample 2

s₁ = ∑(x₁ - x₁')²/n₁,  s₂ = ∑(x₂ - x₂')²/n₂

The test statistic is a variable that is derived from a given data sample and is applied in hypothesis testing. The test statistic measures the available data against the expected value from the null hypothesis

With the given data, we have

t = \frac{0.49-0.29}{\sqrt{\frac{(0.42)^2}{100} +}\frac{(0.29)^2}{100}  }

4 0
3 years ago
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