Solve the equation below 9 = x/18 x= What is X?

When purchased, a beach ball is deflated and packaged flat. The package states that when inflated to its maximum capacity, the r

MrRissso [65]

volume for a sphere (ball) is 4/3 x PI x r^3

r = 6.3, r^3 = 6.3^3 = 250.047

volume = 4/3 x 3.14 x 250.047 = 1046.863

rounded to nearest hundredth = 1046.86 cubic inches

5 0

3 years ago

Question in picture. Thanks!

KonstantinChe [14]

2x^2+4x+2
Carry the 2 over to the 3 behind the 4 and multiply so 2times2 is 4. So i new equation is.
X^2+4x+4
Now you have to find the factors of 4 which is 1,4 and 2,2
So now you have to see which one adds up to 4 so 1+4 is 5 so cross that out and 2+2=4.
So the two number is 2 and 2 now you have to bring down the x^2 and the signs.
(X+2) (X+2) and thats your answer

8 0

3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let $\mu$ = true average percentage of organic matter

So, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics = $\frac{2.481-3}{\frac{1.616}{\sqrt{30} } }$ ~ $t_2_9$

= -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0

3 years ago

A manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs

Hitman42 [59]

Answer:

Step-by-step explanation:

Let X be the solvent present

Then given that the manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs of solvent.

X is N(101.3, 3.68)

a) $P(X>100) = P(Z>\frac{100-101.3}{3.68} =P(Z>-0.35)\\=0.6368$

Probability = 0.6368

b) $P(X>100)= 0.97$ means Z score = 1.88

X=100+1.88(s) =101.3

s=1.3/1.88=0.6914

4 0

3 years ago

Hear is a diagram showing a mixture of red paint and green paint needed for 1 batch of a particular shade of brown

Lelechka [254]

From the diagram above

mixtures of 3 cups of red paints and 2 cups of green paints for 1 batch

this means for every one batch of brown paint

you have 3 cups of red paints and 2 cups of green paints

for three batches

3 red cups + 3 red cups + 3 red cups = 9 reds cup

2 green cups + 2 green cups + 2 green cups = 6 greens cup

to produce three batches you will need the mixture of 9 red cups of paints and 6 green cups of paint

adding to the diagram you will have 6 more red cups and 4 more green cups

4 0

1 year ago