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netineya [11]
3 years ago
13

9th one pls fast as soon as possible

Mathematics
2 answers:
erica [24]3 years ago
4 0
The answer is -1... which is option a. I had a to be -0.69 and b to be -0.31.. l hope it helped??
ladessa [460]3 years ago
4 0

Answer

Value of a3³+b³-3ab = -1

Explanation

Given a+b=-1----(1)

On cubeing both sides of equation (1), we get

(a+b)³ = (-1)³

Call 98%

we know the algebraic identity:

(x+y)³ = x³+y³ + 3xy(x+y)

=> a³ + b³ +3ab(a+b)= -1

=> a³ + b³ +3ab(-1)=-1

/* from (1)*/

=> a³ + b³-3ab = -1

Therefore,

Value of a³+b³-3ab = -1

Please give me brainliest (click on the crown) Thank you :)

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Carl’s square swimming pool has a volume of 1000 cubed feet. What are dimensions of his pool?
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3 years ago
Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w
Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

Z= \frac{(X[bar]-Mu)}{\frac{Sigma}{\sqrt{n}}  } ~N(0;1)

This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.

Z_{\alpha } = Z_{0.01} = -2.334

If Z ≤ -2.334, the decision is to reject the null hypothesis.

If Z > -2.334, the decision is to not reject the null hypothesis.

Using the data of the sample I've calculated the sample mean.

X[bar]= ∑X/n= 1548.61/26= 59.56 cal

Z_{H_0}= \frac{(59.56-60)}{\frac{0.82}{\sqrt{26} } } = -2.736

The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.

At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.

I hope it helps!

6 0
3 years ago
Is 2000 1 tenth of 200?
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Simplify 0.95x-9-1/2x(36-18y)-0.25(-3x+4.8)
iris [78.8K]

Answer: 9xy−16.3x−10.2

Step-by-step explanation:

Distribute:

=0.95x+−9+9xy+−18x+(−0.25)(−3x)+(−0.25)(4.8)

=0.95x+−9+9xy+−18x+0.75x+−1.2

Combine Like Terms:

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