HELP! -1/4 × (-7/9)

What is the value of a?

Burka [1]

Answer:

9

Step-by-step explanation:

3 0

3 years ago

What is the value of 5 in 756 write and draw to explain how you know?

kykrilka [37]

In the attached diagram you can see Place Value Chart that is used to help understand the value of each digit based on the place or position.

For the number 756:

7 has a value of 7 hundreds, or 700;
5 has a value of 5 tens, or 50;
6 has a value of 6 ones, or 6.

Answer: 5 has a value of 5 tens, or 50

8 0

4 years ago

Read 2 more answers

A spherical tank has a radius of 6 yards. It is filled with a liquid the cost $7.15 per cubic yard.

mote1985 [20]

We have two questions, but we need to find the volume of the spherical tank first.
So the formula we will use to find the volume is:
(4/3)πr³=V
Let's fill in the information we have and solve. We will be using 3.14 for pi.
(4/3)(3.14)6³
(4/3)(3.14)(216)
4.1867×216
I'll be using the repeating decimal for 4.1867, I rounded to 7, but there is a repeating 6.
904.32
So the volume is 904.32 yards³ (or 904.78 if using the full pi)
We have to multiply the volume by 7.15.
904.32×7.15
6,465.89
It will cost $6,465.89 to fill the tank with the liquid.

3 0

3 years ago

Read 2 more answers

Let P=x^2+6 PLEASE SHOW WORK!! 15 POINTS

stellarik [79]

Answer:

Step-by-step explanation:

P = x^2 + 6

$(x^2+6)^2 - 21 = 4x^2 + 24\\(x^2+6)^2 - 4x^2 - 45 = 0$

Solving for 4x^2:

P = x^2 +6

P - 6 = x^2

4(P - 6) = 4x^2

Substitution:

$p^2 - 4p - 24 - 45 = 0\\p^2 -4p - 69 = 0$

4 0

4 years ago

Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt

-Dominant- [34]

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is tested if it does not outlast by more than 2 hours, that is, the subtraction is not more than 2:

$H_0: \mu_B - mu_A \leq 2$

At the alternative hypothesis, it is tested if it outlasts by more than 2 hours, that is:

$H_1: \mu_B - \mu_A > 2$

The sample means are: $\mu_A = 8.65, \mu_B = 11.23$

The standard deviations for the samples are $s_A = s_B = 0.67$

Hence, the standard errors are:

$s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934$

The distribution of the difference has mean and standard deviation given by:

$\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58$

$s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 2$ is the value tested at the hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{2.58 - 2}{0.2735}$

$t = 2.12$

The critical value for a right-tailed test, as we are testing if the subtraction is greater than a value, with a 0.05 significance level and 12 + 12 - 2 = 22 df is given by $t^{\ast} = 1.71$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

7 0

3 years ago