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3 years ago

PLEASE HELP ASAP I WILL MARK BRAINLIEST!!!

Mathematics

2 answers:

Setler79 [48]3 years ago

3 0

the answer on number 14. is ×=3 your welcome now please give me brainlest

Mazyrski [523]3 years ago

3 0

13: y = x^2 - 6x + 9

14: x^4 + 2x^3 + x^2 - 3

15: y = -4x^2 + 16x - 12

Hope this helps. (:

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Scores for a civil service exam are normally distributed with a mean of 75 and a standard deviation of 6.5. To be eligible for c

kvasek [131]

Answer:

The lowest score you can earn and still be eligible for employment is 85.6925

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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 75, \sigma = 6.5$

Top 5%.

At least the 100-5 = 95th percentile.

The 95th percentile is X when Z has a pvalue of 0.95. So X when Z = 1.645.

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

$1.645 = \frac{X - 75}{6.5}$

$X - 75 = 1.645*6.5$

$X = 85.6925$

The lowest score you can earn and still be eligible for employment is 85.6925

3 0

4 years ago

Read 2 more answers

How can we change a linear equation ( y = -9.4x+7.4 ) to a quadratic equation?

babunello [35]

Answer:

47x+5y=37 is the answer to your question .

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Scores on a test have a mean of 70 and q3 is 83. the scores have a distribution that is approximately normal. find p90. (you wil

Hunter-Best [27]

<span>Answer: Q3 represents 75%, meaning a z of ~0.67 80 - 70 is 10, so the standard deviations is ~14.9. 10 / 0.67 = 14.9 now find the z that represents a score of 90 90 - 70 is 20 20 / 14.9 = 1.34 from a z-table, a z of 1.34 represents a probability of ~90.99% meaning that there is about a 9.01% chance of getting a 90 or better.</span>

5 0

4 years ago

The points (2,4) and (-2,-4) are plotted on a coordinate plane the equation is y = a * x what is A?

laila [671]

A equals to 2.

<u>Step-by-step explanation:</u>

y = mx + c is the equation of the line and m is the slope and c is the y- intercept.

In the given equation, y = ax , a is the slope.

Slope A can be found by means of the formula, and using the points (2,4) and (-2-4)

slope, A = (y₂ -y₁) / (x₂-x₁)

Plugin the values as,

A = (-4-4) / (-2-2) = (-8) / (-4) = 2

So the value of A is 2.

5 0

3 years ago

A coin is tossed 71 times and 34 heads are observed. Is the coin fair? Use a 98% level confidence interval to base your inferenc

Fofino [41]

Answer:

a) $\hat p=\frac{34}{71}=0.48$ estimated proportion of heads observed.

b) $Se= \sqrt{\frac{0.479(1-0.479)}{71}}=0.0593$

c) $z_{\alpha/2}=2.33$

d) $0.479 - 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.341$

e) $0.479 + 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.617$

f) In order to be fair the coin needs to have included the 0.5 on the interval, and for this case it's included, so we can say that at 2% of significance we can say that the coin is fair.

Step-by-step explanation:

Notation and definitions

$X=34$ number of heads observed.

$n=71$ random sample taken

$\hat p=\frac{34}{71}=0.479$ estimated proportion of heads observed.

$p$ true population proportion of heads observed.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The point estimator for the population proportion of heads is: Answer with two decimal precision.

$\hat p=\frac{34}{71}=0.48$ estimated proportion of heads observed.

The standard error in this estimate is: Answer with four decimal precision.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And the standard error is given by:

$Se= \sqrt{\frac{\hat p(1-\hat p)}{n}}$

$Se= \sqrt{\frac{0.479(1-0.479)}{71}}=0.0593$

The correct table value for the 98% level confidence interval is: Answer with two decimal precision

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

The lower end point of the CI is: Answer with three decimal precision.

And replacing into the confidence interval formula we got:

$0.479 - 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.341$

The upper end point of the CI is: Answer with three decimal precision.

$0.479 + 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.617$

And the 90% confidence interval would be given (0.341;0.617).

Based on the confidence interval, it is plausible that the coin is fair.

In order to be fair the coin needs to have included the 0.5 on the interval, and for this case it's included, so we can say that at 2% of significance we can say that the coin is fair.

4 0

4 years ago