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3 years ago

12

a certain number is both a perfect square and a perfect cube. Will it's square root and its cube root will always be different n

umbers?

1 answer:

Alexus [3.1K]3 years ago

6 0

Answer:

No

Step-by-step explanation:

The number 1 is a square number and a cube number, but its square root and cube root will always be the same(1).

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The sixth-graders at Lester's school got to visit either the science museum or the history museum. 21 students picked the scienc

Lina20 [59]

21 + 28 = 49 students

Ration of history 21 to students 49 = d)21:49

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3 years ago

E^(i.π)+2^(-2+3-1)+21899921030

Anna007 [38]

Answer:

21899921030

Step-by-step explanation:

e^(i.π) = -1

2^(-2+3-1) = 2^0 = 1

Therefor

-1 + 1 + 21899921030 = 21899921030

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2 years ago

Using the distributive property, which expression is equivalent to 6(g + 3)?

const2013 [10]

Step-by-step explanation:

Property: $a(b+c)=ab+ac$

<u />

<u>Use the Distributive Property:</u>

<u /> $6(g+3) =6(g)+6(3)=6g+18$

6g + 18 is an equivalent expression when the distributive property is used.

8 0

2 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

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

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago

The polynomial of degree 4, P ( x ) has a root of multiplicity 2 at x = 4 and roots of multiplicity 1 at x = 0 and x = − 4 . It

Annette [7]

Answer: P(x) = {(x-4)^2} (x) (x+4)

Step-by-step explanation:

Let's start with the multiplicity of 2;

At multiplicity of 2; x=4.

Therefore, x - 4 is a factor of the function P(x).

Since it has a multiplicity of 2, we will rewrite the factor as (x-4)^2

Now for the multiplicity of 1.

At this multiplicity of 1, x= 0 and - 4.

Therefore, the factors are x-0 and x+4

Since multiplicity of 1, the factors remain as they are without any additional root on top.

Therefore, the factors of the polynomial p(x) are (x-4)^2 and x and x+4.

And solution of P(x) in factor form will be: P(x) = {(x-4)^2} (x) (x+4)

7 0

3 years ago