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3 years ago

Which group had the highest percentage of students like Sport.

Mathematics

1 answer:

Vesnalui [34]3 years ago

5 0

The group that had the highest percentage of students that like Sports is group B.

The group that had the less percentage of students that like English is group A.

the percentage of students like Math in group A is 30% and 10% in group B

<h3>What are the percentages?</h3>

A pie chart is a graph that displays information in a circle. The circle is divided into slices which represent a numerical proportion. The sum of angles in a pie chart is 360 degrees

Percentage of students that like sports in:

Group A = [20 / (20 + 15 + 5 + 10)] x 100 = 40%
Group B = [22 / (5 + 4 + 9 + 22)] x 100 = 55%

Percentage of students that like English in:

Group A = [5 / (20 + 15 + 5 + 10)] x 100 = 10%
Group B = [5 / (5 + 4 + 9 + 22)] x 100 = 22.5%

Percentage of students that like Maths in:

Group A = [15 / (20 + 15 + 5 + 10)] x 100 = 30%

Group B = [4 / (5 + 4 + 9 + 22)] x 100 = 10%

To learn more about pie charts, please check: brainly.com/question/26250795

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Step-by-step explanation:

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A small manufacturing firm has 250 employees. Fifty have been employed for less than 5 years and 125 have been with the company

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Answer:

Step-by-step explanation:

Given that a small manufacturing firm has 250 employees. Fifty have been employed for less than 5 years and 125 have been with the company for over 10 years. So remaining 75 are between 5 and 10 years.

Suppose that one employee is selected at random from a list of the employees

A) Probability that the selected employee has been with the firm less than 5 years = $\frac{50}{250 } \\= 0.20$

B) Probability that the selected employee has been with the firm between 5 and 10 years

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C) Probability that the selected employee has been with the firm more than 10 years

= $\frac{125}{250} =0.50$

a) P(A) = 0.2

P(C) = 0.5

P(A or B) = 0.2+0.3 = 0.5

P(A and C) = 0 (since A and C are disjoint)

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3 years ago

Can someone tell me the answer to this?

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Answer:

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At the start of the year, 15 chameleons were introduced into a zoo. The population of chameleons is expected to grow at a rate o

bearhunter [10]

Answer:

option-B

Step-by-step explanation:

We are given

At the start of the year, 15 chameleons were introduced into a zoo

so, $P_0=15$

The population of chameleons is expected to grow at a rate of 41.42% every year

so, r=0.4142

and x represents the number of years since the chameleons were introduced into the zoo

now, we can set equation to find total population

and we get

$P(x)=P_0(1+r)^x$

now, we can plug values

$P(x)=15(1+0.4142)^x$

$P(x)=15(1.4142)^x$

Average rate of change between 2 years and 4 years:

we can use formula

$A_1=\frac{P(4)-P(2)}{4-2}$

now, we can plug values

$A_1=\frac{15(1.4142)^{4}-15(1.4142)^{2}}{4-2}$

$A_1=14.99914$

Average rate of change between 4 years and 6 years:

we can use formula

$A_2=\frac{P(6)-P(4)}{6-4}$

now, we can plug values

$A_2=\frac{15(1.4142)^{6}-15(1.4142)^{4}}{6-4}$

$A_2=29.99770$

Average rate of change between 6 years and 8 years:

we can use formula

$A_3=\frac{P(8)-P(6)}{8-6}$

now, we can plug values

$A_3=\frac{15(1.4142)^{8}-15(1.4142)^{6}}{8-6}$

$A_3=59.99425$

now, we will check each options

option-A:

we can see that

$A_3-A_2=30$

So, this is FALSE

option-B:

$A_1=\frac{1}{2}A_2$

So, this is TRUE

option-C:

This is FALSE

option-D:

we got

$A_1=\frac{1}{2}A_2$

so, this is FALSE

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3 years ago

Show me how to solve 172 times 6

stepan [7]

I would recommend starting by turning 172 into 100+70+2. This makes it much easier to multiply it by 6.

(100+70+2) × 6 =

distributive property

(100×6) + (70×6) + (2×6) =

Simplify

600 + 420 + 12 =

Add it all together

1032

The answer would be 1032.

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3 years ago