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Bogdan [553]
3 years ago
15

A teacher believes that the third homework assignment is a key predictor in how well students will do on the midterm. Let x repr

esent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below. Assume scores are normally distributed.Body frame size is determined by a person's wrist circumference in relation to height. A researcher measures the wrist circumference and height of a random sample of individuals.What is the value of the test statistic to see if the correlation is statistically significant
Mathematics
1 answer:
Darina [25.2K]3 years ago
4 0

Answer:

Hello your question has a disjointed equation attached to it and it is also incomplete attached below is the correct and complete question

A teacher believes that the third homework assignment is a key predictor in how well students will do on the midterm. Let x represent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below

answer :

y-intercept = 25.9047

 slope = 2.8420

Step-by-step explanation:

Determine the slope for the regression equation and y intercept

The regression equation ( gotten using excel ; attached below is the excel sample on how the equation was gotten )

y = 25.9047 + 2.8420x

from the equation gotten above

y-intercept = 25.9047

slope = 2.8420

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The radius of a circle is 2.9 in. Find the circumference to the nearest tenth.
ELEN [110]

Answer:

\bf \large 20 \: inches

Step-by-step explanation:

<h2>Given that :</h2>

  • The radius of a circle is 2.9 in.

<h2>to find :</h2>

  • Find the circumference to the nearest tenth.

<h2>formulas used :</h2>

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where,

  • π = 22/7
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<h2>explanation :</h2>

⟼ c = 2πr

⟼ c = 2 × 22/7 × 2•9 inches

⟼ c = 2 × 3•14 × 2•9 inches

⟼ c = 6•28 × 2•9 inches

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<h2>Round to the nearest tenth :</h2>

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3 0
3 years ago
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st
Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so \mu = 0.5, \sigma = 0.05.

What is the probability that a line width is greater than 0.62 micrometer?

That is P(X > 0.62)

So

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

Z = \frac{0.62 - 0.5}{0.05}

Z = 2.4

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

P(X \leq 0.62) + P(X > 0.62) = 1

P(X > 0.62) = 1 - 0.99180 = 0.0082

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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3 and 2 should be the only roots
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