hello help Plis……..

Write an algebraic expression to represents the verbal expression: the difference between the product of 4 and a number and the

Naily [24]

Answer:

$4n-n^2$

Step-by-step explanation:

I use 'n' as the unknown value, but you can use any letter you want as the variable.

The product of four and a number would be represented by '4 * n'. The ideal way to write this would be '4n', even though '4 * n' is also correct in value.

The square of the number would be n². Squaring a number is multiplying a number by itself.

The phrase 'difference between...' means subtraction.

Putting it together:

"The difference between the product of 4 and a number and the square of the number" would be:

$4n-n^2$

Hope this helps.

4 0

3 years ago

company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of attaining a fleet average of at least 26

podryga [215]

Answer:

$t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435$

$p_v =P(t_{(49)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation

$\bar X=25.02$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =26$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is at least 26 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \geq 26$

Alternative hypothesis: $\mu < 26$

If we analyze the size for the sample is > 30 but 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:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=50-1=49$

Since is a one sided test the p value would be:

$p_v =P(t_{(49)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

3 0

3 years ago

Study polygons PQRS and EFGH in the graph. Which sentence describes a similarity transformation that maps polygon PQRS to polygo

JulijaS [17]

turn EFGH 90° twice in a clockwise Direction

4 0

3 years ago

At Pike Place Fish Market in Seattle, customers can purchase a variety of different types of seafood. One type of seafood sold a

allsm [11]

Answer:

$z=1.28$

And if we solve for a we got

$a=3.6 +1.28*0.8=4.624$

So the value of height that separates the bottom 90% of data from the top 10% is 4.624.

So then the best answer for this case would be:

C. 4.64

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.6,0.8)$

Where $\mu=3.6$ and $\sigma=0.8$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=3.6 +1.28*0.8=4.624$

So the value of height that separates the bottom 90% of data from the top 10% is 4.624.

So then the best answer for this case would be:

C. 4.64

8 0

3 years ago

Approximately 1 in 28 people of Ashkenazi Jewish descent are Tay-Sachs carriers. In randomly sampling one man and one woman from

allsm [11]

Answer:

a) probability that neither is Tay-Sachs carrier is 0.9298

b) the probability that both are carriers is 0.001276

Step-by-step explanation:

Given that;

1 in 28 people of Ashkenazi jewish descent are Tay-Sachs carrier.

so P(Carrier) = 1/28 =

also P( Not a carrier) = 1 - 1/28 = 0.9643

Now After Sampling 1 man and 1 woman from the population;

a)

the probability neither is Tay-Sachs carrier

P( Neither is a Carrier) = 27/28 × 27/28 = 0.9298

Therefore probability that neither is Tay-Sachs carrier is 0.9298

b)

the probability both are carriers

P( Both are Carriers) = 1/28 × 1/28 = 0.001276

Therefore the probability that both are carriers is 0.001276

8 0

3 years ago

￼hello help Plis……..

hello help Plis……..