All categories

2 years ago

A pack of bottel of juice cost 9.09 how much does each bottle cost to the nearest cent

Mathematics

2 answers:

Lapatulllka [165]2 years ago

8 0

Answer:

909 cents

Step-by-step explanation:

There are 100 cents in 1 dollar

9.09 dollars * 100 cents = 909 cents

IRISSAK [1]2 years ago

5 0

How many bottles are there in a pack?
Lets say there are 3
if there are 3 bottles in a pack
and
3 bottles cost 9.09
then 1 bottle would cost 9.09 divided by 3 therefore 1 bottle would cost 3.03

You might be interested in

Five times a number is the same as two times a number added 21

Vera_Pavlovna [14]

Answer:

Our number is 7.

Step-by-step explanation:

We also know that it's 2x + 21 = 5x.

Now we solve.

21 = 5x - 2x

21 = 3x

7 = x

7 0

3 years ago

115.376 as hundreds tens and ones

Gennadij [26K]

Answer:

NAH FAM- I GOTCHU

RATATATA-

YA LIFE IS A LIE~

JFAIOSDHFAIOSDHFSJD~

FAISDEJHFIAJDE~

WOOOOOOOOOOOOOOOOOOOOOOOOOOO

Step-by-step explanation:

5 0

3 years ago

Given C=n+15, determine the value of C when n=4

puteri [66]

Answer: c=19 and n=4

please mark as brainliest if it helped

6 0

3 years ago

Read 2 more answers

Write the number thirty five million thirty thousand

Shtirlitz [24]

35,030,000 is the correct answer, hope this helps.

7 0

3 years ago

Read 2 more answers

Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 c

steposvetlana [31]

Answer:

There is a significant difference between the two proportions.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for difference between population proportions is:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

Compute the sample proportions as follows:

$\hat p_{1}=\frac{80}{250}=0.32\\\\\hat p_{2}=\frac{40}{175}=0.23$

The critical value of <em>z</em> for 90% confidence interval is:

$z_{0.10/2}=z_{0.05}=1.645$

Compute a 90% confidence interval for the difference between the proportions of women in these two fields of engineering as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

$=(0.32-0.23)\pm 1.645\times\sqrt{\frac{0.32(1-0.32)}{250}+\frac{0.23(1-0.23)}{175}}\\\\=0.09\pm 0.0714\\\\=(0.0186, 0.1614)\\\\\approx (0.02, 0.16)$

There will be no difference between the two proportions if the 90% confidence interval consists of 0.

But the 90% confidence interval does not consists of 0.

Thus, there is a significant difference between the two proportions.

4 0

3 years ago