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3241004551 [841]
3 years ago
6

A rectangle has a height 3z and a width of 2x-3 Express the area of the entire rectangle.

Mathematics
1 answer:
viktelen [127]3 years ago
5 0

Answer:

3z(2x - 3)

Step-by-step explanation:

Area of a rectangle, a = height * width

height, h = 3z

width, w = 2x - 3

a = 3z * (2x -3)

expand

a = 6xz - 9z

a = 3z(2x - 3)

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Mike spent half of his allowance going to the movies. He washed the family cat and earned $7. What is his weekly allowance if he
Natasha_Volkova [10]
16 - 7 is 9
9 x 2 is 18
His weekly allowance is 18
4 0
3 years ago
PLEASE HELP ME !<br> (2,7); 3x-y =5<br> show all steps !
Zigmanuir [339]

Answer:

i dont think you've got the right (x,y) pair....

Step-by-step explanation:

3x - y = 5

3(2) - 7 = 5

6 - 7 = 5

those are not the correct coordinates for the equation

3 0
3 years ago
When 5 times a number is added to 45 the answer is the sm of 85 and 140 what is the number
zimovet [89]

Hello.

First, let the number be r.

5 times r is written like this:

\rm{5r}

Now, add 45:

\mathrm{5r+45}

Now, the sum of 5r+45 is equal to the sum of 85 and 140:

\mathrm{5r+45=85+140}

In order to solve this equation, we need to add 85 and 140:

\mathrm{5r+45=225}

Now, subtract 45 from both sides:

\mathrm{5r+45-45=225-45}

\mathrm{5r=225-45}

\mathrm{5r=180}

The last step is to divide both sides by 5:

\mathrm{r=36}

Therefore, the answer is

\rm{r=36}

I hope it helps.

Have a nice day.

\boxed{imperturbability}

8 0
3 years ago
If the 12-pack of water bottles at Kroger costs $4.60 then How much would one of those bottles of water cost?
xeze [42]

Answer:

Step-by-step explanation:

If the 12-pack of water bottles at Kroger costs $4.60 then How much would one of those bottles of water cost?

1. Divide the cost by the number of bottles.

4.60 ÷ 12 = $0.38/bottle

8 0
3 years ago
The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed
Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

      =\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911

Result is not significant.

3.

(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378

Result is not significant.

4.

(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349

Result is not significant.

5.

(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773

Result is not significant.

6.

(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.

   

8 0
3 years ago
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