Use benchmarks to estimate the sum 11/15 + 1/8

Please answer with explanation of how you got the answer !!

Irina18 [472]

Answer:

−24

Step-by-step explanation:

Since there is only one negative, this means when all the values are multiplied together the product will be negative. Just multiply all three together to get −24.

(3)(−4)(2)

(−12)(2)

−24

It doesn't matter which order you multiply in in this case.

7 0

3 years ago

Find m21. 103°. 13° 77°. 26°

barxatty [35]

If you mean find the one that has the number 1 on it your answer is 13.
90+77+x=180
167+x=180
X=13

8 0

2 years ago

Write 4.135 as a fraction

Nuetrik [128]

Answer:

Step-by-step explanation:

827/200

8 0

3 years ago

Read 2 more answers

15Points!! Factor completely. 12x² +28x

Zigmanuir [339]

Answer:

4x(3x+7)

Step-by-step explanation:

$12x^2+28x\\12xx+28x\\\\4\cdot \:3xx+4\cdot \:7x\\4x\left(3x+7\right)$

4 0

2 years ago

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in

erica [24]

Answer:

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$n=10$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.0217$ represent the sample variance

$\sigma^2_0 =0.12^2= 0.0144$ represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 0.0144$

Alternative hypothesis: $\sigma^2 \neq 0.0144$

The statistic can be calculated like this;

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

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

7 0

3 years ago