-50 added to three times an unknown number equals -11? What is the solution to the sentence shown?

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question.

inn [45]

Answer:

Step-by-step explanation:

(a) MEAN = £fx/ fx (The sum of the all jerseys number/ number of jerseys)

= 680/11

= 61.8

(b) MEDIAN = The middle number when all the items are arranged either in ascending order or descending order

= 6, 14, 50, 51, 54, 68, 77, 78, 89, 94, 99

= 68

(Strike out the numbers, the middle number left is the median)

(c) MODE= (Mode implies the number with the highest frequency)

From those items, THERE IS NO MODE.

All the items (numbers) appeared once

(d) MID-RANGE:

This is the sum of the least valued number and the greatest valued number and dividee by two

Mid-range = (99 + 6) / 2

= 105/2

= 52.5

(e). The results tells us that:

The mean and median give two different interpretations of the average (or typical) jersey number, while the midrange shows the spread of possible jersey numbers

Also, pls note that the median implies the measure of the middle of the sets of jerseys number. And the mean tells us the distribution of the sets of jerseys number.

Hope this helped!

3 0

3 years ago

Is this problem convergent or divergent?

Svetllana [295]

Answer:

Its divergent

Step-by-step explanation:

HOPE YOU LIKED IT

6 0

2 years ago

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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 mean of a population is 74 and the standard deviation is 15.

This means that $\mu = 74, \sigma = 15$

Question a:

Sample of 36 means that $n = 36, s = \frac{15}{\sqrt{36}} = 2.5$

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

$Z = \frac{78 - 74}{2.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that $n = 150, s = \frac{15}{\sqrt{150}} = 1.2247$

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

$Z = \frac{77 - 74}{1.2274}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 71

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

$Z = \frac{71 - 74}{1.2274}$

$Z = -2.45$

$Z = -2.45$ has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that $n = 219, s = \frac{15}{\sqrt{219}} = 1.0136$

This probability is the pvalue of Z when X = 74.2. So

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

$Z = \frac{74.2 - 74}{1.0136}$

$Z = 0.2$

$Z = 0.2$ has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0

3 years ago

Polynomial equation when zeros are -4, -2, 1, 4

jek_recluse [69]

<h2>The required poynomial equation is $x^{4}+x^{3}-18x^{2}-16x+32$ .</h2>

Step-by-step explanation:

Given,

The zeros of the polynomial are - 4, - 2, 1 and 4

To find, the polynomial equation = ?

We know that,

The polynomial equation = (x - A)(x - B)(x - C)(x - D)

= (x + 4)(x + 2)(x - 1)(x - 4)

= [(x + 4)(x - 4)] [(x + 2)(x - 1)]

= ( $x^{2}-16$ )( $x^2-x+2x-2$ )

= ( $x^{2}-16$ )( $x^2+x-2$ )

= $x^{4}+x^{3}-2x^{2}-16x^{2}-16x+32$

= $x^{4}+x^{3}-18x^{2}-16x+32$

Thus, the required poynomial equation is $x^{4}+x^{3}-18x^{2}-16x+32$ .

4 0

2 years ago

The segment in this diagram that divides AB into two congruent parts is called a(n): A. angle bisector. B. parallel line C. perp

melamori03 [73]

C perpendicular bisector.

7 0

2 years ago