If 5 slices of pizza cost 5.50 how much do 2 slices cost 10 half a slice?

A baby elephant that is 4 ft tall casts a

Kazeer [188]

Answer:

15 ft

Step-by-step explanation:

10/4

=2.5

6*2.5

=15 ft

PLS GIVE BRAINLIEST

7 0

3 years ago

Read 2 more answers

Explain how you can use base-ten blocks or as a decimal models to find 3.15 divided by 3. Include pictures to support your expla

AlladinOne [14]

You have to turn the number or ratio and simplified the number and put it in to the chart like the norm number

.

5 0

3 years ago

Juan ordered 20 pizzas for a party. 45% of the pizzas have 8 slices each. The remaining 55% of the pizzas have 12 slices each. C

Mama L [17]

Total number of pizzas Juan ordered $20$

Since, $45\%$ of pizzas have $8$ slices.

Therefore, number of pizzas with $8$ slices are: $45\%\times(20)$

$=\frac{45}{100} \times(20)$

$=\frac{45}{10} \times(2)$

$=\frac{90}{10}$

$=9$ pizzas

Also, since, $55\%$ of pizzas have $12$ slices.

Therefore, number of pizzas with $12$ slices are: $55\%\times(20)$

$=\frac{55}{100} \times(20)$

$=\frac{55}{10} \times(2)$

$=\frac{110}{10}$

$=11$ pizzas

Now, there are $9$ pizzas with $8$ slices and $11$ pizzas with $12$ slices.

Therefore, total number of slices of pizza are: $=(9)(8)+(11)(12)$

$=72+132$

$=204$ slices

5 0

3 years ago

Suppose a local manufacturing company claims their production line has a variance of less than 9.0. A quality control engineer d

Alexxx [7]

Answer:

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

$p_v =P(\chi^2$

If we compare the p value and the significance level provided we see that $p_v$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=35$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s =2.12$ represent the sample deviation obtained

$\sigma_0 =3$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is lower than 9 and the deviation lower than 3, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 9$

Alternative hypothesis: $\sigma^2$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case df= n-1= 35-1=34. And since is a left tailed test the p value would be given by:

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(16.979,34,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

6 0

3 years ago

Two less than thrice a number is 2 more than two third of

harkovskaia [24]

Step-by-step explanation:

Let the two numbers be x and y.

$\therefore \: 3x - 2 = \frac{2}{3} y + 2 \\ \\ \therefore \: 3x - \frac{2}{3} y = 2 + 2\\ \\ \therefore \: 3x - \frac{2}{3} y = 4 \\ \\ \therefore \: 9x - 2y = 12$

8 0

3 years ago