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4 years ago

HELP!!!!!!!! The histogram below shows the estimated monthly salaries of company employees with different years of experience.

Mathematics

2 answers:

Rudiy274 years ago

7 0

The correct answer is A. The scale on the y-axis overemphasizes the difference in salaries between different experience levels.

Likurg_2 [28]4 years ago

5 0

Answer:
_________________________________________________________
The "scale" used on the "x-axis" for the graph shown (i.e. "Years of Experience") does not fully account for the "estimated monthly salaries" (scale on the "y-axis") AMONG different experience levels .
_________________________________________________________

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Veronika [31]

Answer:

99% of the sample means will fall between 0.93288 and 0.94112.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The true mean is .9370 with a standard deviation of 0.0090

This means that $\mu = 0.9370, \sigma = 0.0090$

Sample of 32:

This means that $n = 32, s = \frac{0.009}{32} = 0.0016$

Within what interval will 99 percent of the sample means fall?

Between the 50 - (99/2) = 0.5th percentile and the 50 + (99/2) = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

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

By the Central Limit Theorem

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

$-2.575 = \frac{X - 0.9370}{0.0016}$

$X - 0.9370 = -2.575*0.0016$

$X = 0.93288$

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

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

$2.575 = \frac{X - 0.9370}{0.0016}$

$X - 0.9370 = 2.575*0.0016$

$X = 0.94112$

99% of the sample means will fall between 0.93288 and 0.94112.

6 0

3 years ago

The equation p(t) = 1.e represents a

zvonat [6]

Answer:

(b), (d) and (e)

Step-by-step explanation:

Given

$p(t) = 1 * e^t$

See attachment for $y = p(t)$

Required

Select true statements from the given options

(a) $\ln(30)$ = days the bacteria reaches 30000

We have:

$p(t) = 1 * e^t$

In this case:

$t = \ln(30)$ and $p(t) = 30000$

So, we have:

$30000 = 1 * e^{\ln(30)}$

$30000 = e^{\ln(30)}$

Using a calculator, we have:

$e^{\ln(30)} = 30$

So:

$30000 = 30$

The above equation is false.

(a) is not true

(b) The graph shows that $\ln(20) \approx 3$

We have:

$p(t) = 1 * e^t$

Let t = 3

So;

$p(3) = 1 * e^3$

From the graph, $p(3) = 20$

So:

$20 = 1 * e^3$

$20 = e^3$

Take natural logarithm of both sides

$\ln(20) = \ln(e^3)$

This gives:

$\ln(20) = 3$

(b) is true

(c) $\ln(t) = y$ is the logarithm form of $y = e^t$

We have:

$y = e^t$

Take natural logarithm of both sides

$\ln(y) = \ln(e^t)$

This gives:

$\ln(y) = t$

$\ln(y) = t$ $\ne$ $\ln(t) = y$

(c) is false

(d) $e^4 > 50$ and $\ln(50) < 4$

From the graph, we have:

$e^4 = 54$ --- rough readings

This implies that:

$e^4 > 50$ is true

Because $54 > 50$

Take natural logarithm of both sides

$\ln(54) > \ln(50)$

Rewrite as:

$\ln(50) < \ln(54)$

We have:

$e^4 = 54$

Take natural logarithm of both sides

$\ln(e^4) = \ln(54)$

$4 = \ln(54)$

$\ln(54) = 4$

Substitute $\ln(54) = 4$ in $\ln(50) < \ln(54)$

$\ln(50) < 4$

(d) is true

(e) The graph shows that $10 \approx \ln(2.3)$

We have:

$p(t) = 1 * e^t$

Let t = 2.3

So;

$p(2.3) = 1 * e^{2.3}$

From the graph,

$p(2.3) = 10$ ---- rough readings

So:

$10 = 1 * e^{2.3}$

$10 = e^{2.3}$

Take natural logarithm of both sides

$\ln(10) = \ln(e^{2.3})$

This gives:

$\ln(10) = 2.3$

(e) is true

4 0

3 years ago

How to write an equation in standard form

densk [106]

There ya go hope it helps

6 0

4 years ago

What is the value of s in the equation 3 r equals 10 plus 5 s, when r equals 10?

Artemon [7]

Answer:

s = 4

Step-by-step explanation:

3r = 10 + 5s

3×10 = 10 + 5s

5s = 30 - 10

5s = 20

s = 20 : 5

s = 4

7 0

3 years ago

Read 2 more answers

Substitute x with -2

aleksley [76]

Answer:

-105

Step-by-step explanation:

Substituting is like "plug and chug", so we can plug in -2 into the expression and get $(13 - 4(-2))((-2)^2+5(-2)+1)$ , which simplifies into 21*-5, or -105. $\blacksquare$

7 0

4 years ago

Read 2 more answers