All categories

3 years ago

Use the grid to create a model to solve the percent problem. 12 is 60% of what number? Enter your answer in the box

Mathematics

2 answers:

viva [34]3 years ago

5 0

Answer: 12 is 60 % of 20.

Step-by-step explanation:

Let us take a 10 box grid,

In which every box shows 10 %, ( Shown below )

Hence, the whole grid shows total 100% .

In the grid,

$60\% = 12$

$1\% = \frac{12}{60}$

$1\%=\frac{1}{5}$

$100\% = \frac{100}{5}=20$

Hence, 12 is 60 % of 20.

Elenna [48]3 years ago

3 0

Is means equal to and of means multiply. So....your equation would be 12=60%x. Now since u can't multiply by a percent, u gotta do 12=.6x. Now divide by .6 so that it cancels out. Your answer is 20. Hope this helps!

You might be interested in

At a large midwestern university, a simple random sample of 100 entering freshmen in 1993 found that 20 of the sampled freshmen

nasty-shy [4]

Answer:

B. Z=1.98

Step-by-step explanation:

1) Data given and notation

$X_{1}=20$ represent the number of sampled freshmen finished in the bottom third of their high school class in 1993

$X_{2}=10$ represent the number of sampled freshmen finished in the bottom third of their high school class in 1997

$n_{1}=100$ sample 1 selected

$n_{2}=100$ sample 2 selected

$p_{1}=\frac{20}{100}=0.20$ represent the proportion of sampled freshmen finished in the bottom third of their high school class in 1993

$p_{2}=\frac{10}{100}=0.0023$ represent the proportion ofsampled freshmen finished in the bottom third of their high school class in 1997

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion graduated in the bottom third of their high school class is higher for 1993 than 1997 , the system of hypothesis would be:

Null hypothesis: $p_{1} \leq p_{2}$

Alternative hypothesis: $p_{1} > p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{20+10}{100+100}=0.15$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.20-0.10}{\sqrt{0.15(1-0.15)(\frac{1}{100}+\frac{1}{100})}}=1.98$

4) Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

Since is a one side test the p value would be:

$p_v =P(Z>1.98)=0.024$

If we compare the p value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion graduated in the bottom third of their high school classes in 1993 is not significant higher than the proportion in 1997.