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3 years ago

Evaluate the expression when y=-6. y² +8744

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

8 0

Answer:

8780.

Step-by-step explanation:

-6 times -6 is 36 + 8744

oee [108]3 years ago

6 0

Answer:

8780

Step-by-step explanation:

We simply substitute "-6" in for "y"

$y^2 + 8744\\\\(-6)^2 +8744\\\\36 + 8744\\\\8780$

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If you get 12 how many do you have to get wrong to get an 58%

VladimirAG [237]

Answer:

I'm gonna take some liberty. I assume you mean if there is 12 questions. if so that means you would need to get 5 wrong.

Step-by-step explanation:

Here are step-by-step instructions showing you how we calculated 7 out of 12 as a percentage: The first step is to divide 7 by 12 to get the answer in decimal form: 7 / 12 = 0.5833. Then, we multiplied the answer from the first step by one hundred to get the answer as a percentage: 0.5833 * 100 = 58.33%

3 0

3 years ago

A researcher examines 27 sedimentary samples for bromide concentration. The mean bromide concentration for the sample data is 0.

FrozenT [24]

Answer:

Critical value: $z = 1.645$

The 90% confidence interval for the population mean bromide concentration is (0.376, 0.39).

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$ . This value of z is the critical value

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645*\frac{0.0209}{\sqrt{27}} = 0.0066$

The lower end of the interval is the mean subtracted by M. So it is 0.383 - 0.0066 = 0.376 cc/cubic meter

The upper end of the interval is the mean added to M. So it is 0.383 + 0.0066 = 0.39 cc/cubic meter

The 90% confidence interval for the population mean bromide concentration is (0.376, 0.39).

6 0

3 years ago

A certain brand of refrigerator has a useful life that is normally distributed with mean 10 years and standard deviation 3 years

juin [17]

Answer: 0.55567

Step-by-step explanation:

Given : A certain brand of refrigerator has a useful life that is normally distributed with mean 10 years and standard deviation 3 years. The useful lives of these refrigerators are independent.

i.e.

$\mu=10\text{ years}\\\sigma=3\text{ years}$

Let $X_1$ and $X_2$ are two randomly selected refrigerator's life whose sum will exceed third selected refrigerator $X_3$ .

So that, $X =X_1+X_2-1.9\times X_3$

Mean $=E(X_1)+E(X_2)-E(X_3)=10+10-1.9\times10 =1$

Standard deviation = $\sqrt{(Var(X_1)+Var(X_1)+Var(X_1))}$

$=\sqrt{(3^2+3^2+(1.9\times3)^2)}$

$=7.1056315694$

Z-score : $z=\dfrac{X-E[x]}{\sqrt{Var[x]}}=\dfrac{0-1}{7.0156315694}\approx-0.14$

Now , The probability that the total useful life of two(i..e n=2) randomly selected refrigerators will exceed 1.9 times the useful life of a third randomly selected refrigerator would be :-

$P(Z>-0.14)=P(Z$