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2 years ago

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Sabrina wants to have an average of at least 90 on her quizzes. If she took three quizzes andearned a 95, 86 and 82, what is the

lowest grade she has to earn on the fourth quiz

1 answer:

Marrrta [24]2 years ago

3 0

Answer:

97

Step-by-step explanation:

<u>97 </u>+ 95 + 86 + 82 = 360

360 / 4 = <em>90</em>

<em />

i hope this helps :))

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A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

$\sigma_x = 1.825$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.825$

$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 3.577$

$Lower = 46.423$

$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

<em>So, the 95% confidence interval is (46.4, 53.6)</em>

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

$n = 90$ $E = 2.06584$

<em>Notice that the margin of error decreases when the sample size increases.</em>

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Given m∠JML = 80 and m∠KML = 33, what is m∠JMK

Harrizon [31]

Answer:

<h2>47°</h2>

Step-by-step explanation:

Given m∠JML = 80 and m∠KML = 33, to calculate the value of angle m∠JMK, we will apply the formula since all the angles are all acting at point M.

m∠JML = m∠JMK+m∠KML

Substituting the values given into the equation;

80 = m∠JMK + 33

subtract 33 from both sides of the equation

m∠JMK = 80-33

m∠JMK = 47°

<em>Hence the value of angle m∠JMK is 47°</em>

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It will take her three days

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adelina 88 [10]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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