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17 days ago

Laila wants to create a data display to clearly show the median salary, the highest salary,

Mathematics

2 answers:

lawyer [4K]17 days ago

7 0

Answer:

A box plot

Step-by-step explanation:

Assuming there are 685 employees in her organization

If she opts for a dot plot, she would need to depict many dots, totaling 685, along with a broad range of values on the x-axis.

=> Thus, it's the most suitable option for her.

Should she select a histogram, it implies her desire to categorize numbers into intervals to uncover the frequency distribution pattern within continuous data.

I hope this information serves you well.

</pchoosing>

Zina [3.9K]17 days ago

5 0

what mine indicates. a dot plot: a box plot: a histogram: therefore, Laila should create a: :to exhibit the data. these responses do not make sense here.

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What is the answer? The glee club has 120 cupcakes to sell. they have decided to arrange the cupcakes in the shape of a rectangl

PIT_PIT [3919]

<span>Determine the configuration of columns and rows for the rectangular arrangement of 120 cupcakes.
=> There must be an even number of rows and an odd number of columns.
=> 120 = 2 x 2 x 2 x 15
=> 120 = 8 x 15
=> 120 = 120
Consequently, the glee club should organize the cupcakes in 8 rows and 15 columns.
This totals up to 120 cupcakes altogether.

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8 0

9 days ago

Mike earned x dollars the first week of his new job. He earned 5% more the second week than the first week. Which expression rep

zzz [4022]

The expression indicates that the total earnings of Mike over the two weeks amount to 2.05x dollars

Solution:

It is given that Mike earned x dollars during the first week at his new job,

While in the second week, he earned 5% more than the earnings of the first week.

To determine: Total earnings over the two weeks

From the information provided,

Earnings in the first week = "x" dollars

Earnings in the second week = 5% more than first week earnings

Thus,

Earnings in the second week = x + 5% of x

$\rightarrow x + 5 \% \times x\\\\\rightarrow x + \frac{5}{100} \times x\\\\\rightarrow x + 0.05x = 1.05x$

Therefore, earnings in the second week equal to 1.05x dollars

The total earnings over the two weeks:

Total earnings = Earnings from first week + Earnings from second week

$Total\ Amount = x + 1.05x = 2.05x$

Thus, the expression for the total earnings of Mike for both weeks is 2.05x dollars

5 0

4 days ago

Which are the solutions of x2 = –5x + 8? StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFract

tester [3916]

Answer:

$x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Detailed solution:

Given:

The problem to solve is:

$x^2=-5x+8$

Convert the equation into the standard quadratic form $ax^2+bx +c =0$ , where $a,\ b,\ and\ c$ represent constants.

So, by adding $5x-8$ to both sides, we get:

$x^2+5x-8=0$

Note that $a=1,b=5,c=-8$ .

The roots of this quadratic are found by applying the quadratic formula given as:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=1,b=5,c=-8$ into the formula and calculate for $x$ .

$x=\frac{-5\pm \sqrt{5^2-4(1)(-8)}}{2(1)}\\x=\frac{-5\pm \sqrt{25+32}}{2}\\x=\frac{-5\pm \sqrt{57}}{2}\\\\\\\therefore x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Hence, the roots are:

$x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

4 0

14 days ago

Read 2 more answers

Greg builds a new pond which has a volume of 7.35m3, it is 4.2m long and 50cm deep, what is the width of the pond

AnnZ [3877]

The width of the pond is 3.5 meters.

6 0

17 days ago

Mike runs for the president of the student government and is interested to know whether the proportion of the student body in fa

zzz [4022]

Answer:

We conclude that less than or equal to 50% of the student body supports him significantly.

Step-by-step explanation:

Mike, while campaigning for the student government presidency, is keen to determine if more than 50% of the student body supports him.

A random sample of 100 students was surveyed, with 55 expressing support for Mike.

Let p = the proportion of students backing Mike.

Thus, Null Hypothesis, $H_0$ : p $\leq$ 50% {indicating that the proportion of supporters among the student body is significantly less than or equal to 50%}

Alternate Hypothesis, $H_A$ : p > 50% {indicating that the proportion of supporters among the student body is significantly more than 50%}

The test statistics to be utilized here One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = the sampled proportion in favor of Mike = $\frac{55}{100}$ = 0.55

n = number of sampled students = 100

Therefore, test statistics = $\frac{0.55-0.50}{\sqrt{\frac{0.55(1-0.55)}{100} } }$

= 1.01

The z test statistic is 1.01.

At a significance level of 0.05, the z table provides a critical value of 1.645 for a right-tailed test.

Given that our test statistic falls below the critical value of z (1.01 < 1.645), we lack sufficient evidence to reject the null hypothesis as it remains outside the rejection region, leading to failure to reject our null hypothesis.

As a result, we conclude that less than or equal to 50% of the student body is in favor of him or the proportion supporting Mike is not significantly greater than 50%.

4 0

2 days ago