A​ person's blood pressure is monitored by taking 8 readings daily. The probability distribution of his reading had a mean of 129 and a standard deviation of 7. a. Each observation behaves as a random sample. Find the mean and the standard deviation of the sampling distribution of the sample mean for the eight observations each day. b. Suppose that the probability distribution of his blood pressure reading is normal. What is the shape of the sampling​ distribution? c. Refer to​ (b). Find the probability that the sample mean exceeds 140.

Answers

Answer 1
Answer:

Answer:

a) \mu_(\bar X)= 129

\sigma_(\bar X)= (7)/(√(8))=2.475

b) If the distribution is normal then the sampling distribution would be bell shaped and normal

c) P(\bar X >140)

And we can use the z score formula given by:

z =(\bar X -\mu)/(\sigma_(\bar X))

And replacing we got:

z =(140-129)/(2.475)= 4.44

And then we can find the probability using the complement rule and the normal standard distribution or excel and we got:

P(\bar X>140) = P(Z>4.44) =1-P(z<4.44) \approx 0

Step-by-step explanation:

For this case we have the following info:

n=8 represent the sample size

\bar X=129 the sample mean

s=7 the standard deviation

Part a

If we assume that the distirbution is bell shaped then we can find the parameters like this:

\mu_(\bar X)= 129

\sigma_(\bar X)= (7)/(√(8))=2.475

Part b

If the distribution is normal then the sampling distribution would be bell shaped and normal

Part c

We want this probability:

P(\bar X >140)

And we can use the z score formula given by:

z =(\bar X -\mu)/(\sigma_(\bar X))

And replacing we got:

z =(140-129)/(2.475)= 4.44

And then we can find the probability using the complement rule and the normal standard distribution or excel and we got:

P(\bar X>140) = P(Z>4.44) =1-P(z<4.44) \approx 0


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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn(x) ? 0.]f(x) = 10/x , a= -2f(x) = \sum_{n=0}^{\infty } ______Find the associated radius of convergence R.R = ______

Answers

Rewrite f as

f(x)=\frac{10}x=-\frac5{1-\frac{x+2}2}

and recall that for |x|<1, we have

\displaystyle\frac1{1-x}=\sum_(n=0)^\infty x^n

so that for \left|\frac{x+2}2\right|<1, or |x+2|<2,

f(x)=-5\displaystyle\sum_(n=0)^\infty\left(\frac{x+2}2\right)^n

Then the radius of convergence is 2.

Final answer:

The Taylor series for the function f(x) = 10/x, centered at a = -2, is given by the formula  ∑(10(-1)^n*n!(x + 2)^n)/n! from n=0 to ∞. The radius of convergence (R) for the series is ∞, which means the series converges for all real numbers x.

Explanation:

Given the function f(x) = 10/x, we're asked to find the Taylor series centered at a = -2. A Taylor series of a function is a series representation which can be found using the formula f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + .... For f(x) = 10/x, the Taylor series centered at a = -2 will be ∑(10(-1)^n*n!(x + 2)^n)/n! from n=0 to ∞. The radius of convergence R is determined by the limit as n approaches infinity of the absolute value of the ratio of the nth term and the (n+1)th term. This results in R = ∞, indicating the series converges for all real numbers x.

Learn more about Taylor series here:

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Which decimal number is equivalent to 7/10

Answers

Answer:

0.7

Step-by-step explanation:

Answer:

0.7

Step-by-step explanation:

7/10 can also be written as 7÷10

  • 7÷10=0.7

A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of type A (the original paint) and 9 cans of type B (the modified paint) were selected and applied to similar surfaces. The drying times, in hours, were recorded. The following 98% confidence interval was obtained for μ1 - μ2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B: 4.90 hrs < μ1 - μ2 < 17.50 hrs
What does the confidence interval suggest about the population means?

A. The confidence interval includes 0 which suggests that the two population means might be equal. There​ doesn't appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.
B. The confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.
C. The confidence interval includes only positive values which suggests that the two population means might be equal. There​ doesn't appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.
D. The confidence interval includes only positive values which suggests that the mean drying time for paint type A is smaller than the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.

Answers

This question is not complete, I got the complete one from google as below:

A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of type A (the original paint) and 9 cans of type B (the modified paint) were selected and applied to similar surfaces. The drying times, in hours, were recorded.

The summary statistics are as follows.

Type A                                   Type B

x1 = 76.3 hrs                       x2 = 65.1 hrs

s1 = 4.5 hrs                          s2 = 5.1 hrs

n1 = 11                                  n2 = 9

The following 98% confidence interval was obtained for μ1 - μ2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B:

4.90 hrs < μ1 - μ2 < 17.50 hrs

What does the confidence interval suggest about the population means?

A. The confidence interval includes 0 which suggests that the two population means might be equal. There​ doesn't appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.

B. The confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.

C. The confidence interval includes only positive values which suggests that the two population means might be equal. There​ doesn't appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.

D. The confidence interval includes only positive values which suggests that the mean drying time for paint type A is smaller than the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.

Answer:

Option B is correct - the confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.

Step-by-step explanation:

The 98% confidence interval for the difference in mean drying times of the two types of paints is (4.90, 17.50). This implies that Type A takes between 4.90 and 17.50 hours more to dry than type B paint.

Thus, option B is correct - the confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.

Let Upper C left-parenthesis q right-parenthesis represent the cost, Upper R left-parenthesis q right-parenthesis the revenue, and pi left-parenthesis q right-parenthesis the total profit, in dollars, of producing q items.(a) If Upper C prime left-parenthesis 50 right-parenthesis equals 75 and Upper R prime left-parenthesis 50 right-parenthesis equals 88, approximately how much profit is earned by the 51 Superscript st item?The profit earned from the 51 Superscript st item will be approximately_______ dollar-sign.(b) If Upper C prime left-parenthesis 90 right-parenthesis equals 71 and Upper R prime left-parenthesis 90 right-parenthesis equals 67, approximately how much profit is earned by the 91 Superscript st item?The profit earned from the 91 Superscript st item will be approximately______ dollar-sign

Answers

Answer:

(a)$13

(b) Loss of $4

Step-by-step explanation:

C(q) represents Cost of producing q units.

R(q) represents Revenue generated from q units.

P(q) represents Total Profit made from producing q units.

Marginal analysis is concerned with estimating the effect on quantities such as cost, revenue, and profit when the level of production is changed by a unit amount. For example, if C(q) is the cost of producing q units of a certain commodity, then the marginal cost, MC(q), is the additional cost of producing one more unit and is given by the difference

MC(q) = C(q + 1) − C(q).

Using the estimation

C'(q)≈[TeX]\frac{C(q+1)-C(q)}{(q+1)-q}[/TeX]=C(q+1)-C(q)

We find out that MC(q)=C'(q)

We can therefore compute the marginal cost by the derivative C'(q).

This also holds for Revenue, R(q) and Profit, P(q).

(a) If C'(50)=75 and R'(50)=88

51st item.

P'(50)=R'(50)-C'(50)

=88-75=$13

The profit earned from the 51st item will be approximately $13.

(b) If C'(90)=71 and R'(90)=67, approximately how much profit is earned by the 91st item.

P'(90)=R'(90)-C'(90)

=67-71= -$4

The profit earned from the 91 st item will be approximately -$4.

There was a loss of $4.

Help pls is my homework helppppppppp
15 points​

Answers

1.B. 2.A 3.A 4.D 5.B

Evaluate the expression 2x2 - yl yl + 3xº for x = 4 and y = 7​

Answers

Answer:

replace the given values of x and y

Step-by-step explanation:

hope it helped!!!