Number line question PLEASE HELP

velikii [3]

This does not make sense,May you put it a little more clear?

5 0

3 years ago

A taxi company is trying to decide whether to purchase brand A or brand B tires for its fleet of taxis. To estimate the differen

gavmur [86]

Answer:

Step-by-step explanation:

The question is incomplete. The complete question is:

A taxi company is trying to decide whether to purchase brand A or brand B tires for its fleet of taxis. To estimate the difference in the two brands, an experiment is conducted using 8 of each brand, assigned at random to the left and right rear wheels of 8 taxis. The tires are run until they wear out and the distances, in kilometers, are recorded in the accompanying data set. Find a 95% confidence interval for Hx - Hy. Assume that the differences of the distances are approximately normally distributed Data Set Let, Hx be the population mean for brand A and let Hy be the population mean for brand B. The confidence interval is -42- (Round to one decimal place as needed.) Taxi Brand A 37,700 46,800 36,500 40,600 43.600 31,300 37,600 43.800 Brand B 39,800 47,500 37.100 39,200 42,900 36,000 38,700 45,000

Solution:

For brand A,

Mean,x1 = (37700 + 46800 + 36500 + 40600 + 43600 + 31300 + 37600 + 43800)/8 = 39737.5

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (37700 - 39737.5)^2 + (46800 - 39737.5)^2 + (36500 - 39737.5)^2 + (40600 - 39737.5)^2 + (43600 - 39737.5)^2 + (31300 - 39737.5)^2 + (37600 - 39737.5)^2 + (43800 - 39737.5)^2 = 172438750

Standard deviation = √(172438750/8

s1 = 4642.72

For brand B,

Mean,x2 = (39800 + 47500 + 37100 + 39200 + 42900 + 36000 + 38700 + 45000)/8 = 40775

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (39800 - 40775)^2 + (47500 - 40775)^2 + (37100 - 40775)^2 + (39200 - 40775)^2 + (42900 - 40775)^2 + (36000 - 40775)^2 + (38700 - 40775)^2 + (45000 - 40775)^2 = 111635000

Standard deviation = √(111635000/8

s2 = 3735.56

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 95% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (8 - 1) + (8 - 1) = 14

z = 2.145

x1 - x2 = 39737.5 - 40775 = -1037.5

z√(s1²/n1 + s2²/n2) = 2.145√(4642.72²/8 + 3735.56²/8)

= 4519.1

95% Confidence interval = - 1037.5 ± 4519.1

3 0

3 years ago

What is the value of 2 in 5,670,249,114? And describe 2 ways to find value?

kirill115 [55]

There are two ways in finding the place value of 2 in 5,670,249,114
First by breaking down the numbers:
5,670,249,114 = 5,000,000,000 + 600,000,000 + 70,000,000 + 200,000 + 40, 000 + 9,000 + 100 + 10 + 4 which means the Value of 2 is 200, 000<span>2. Second, by using the ones words:
5,670,249,114 = 5 billion + 6 hundred million + 70 ten million + 2 hundred thousand + 4 ten thousand + 9 thousands + 1 hundreds + 1 tens + 4 ones which means the Value of 2 is 2 hundred thousand

</span>

4 0

3 years ago

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

3 years ago

The average cost of printing a book in a publishing company is c(x) =, <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B5.5x%2Bk%7

Bogdan [553]

To find K, we must use the fact that
on the day 200 books were printed
the average cost was $ 9 per book.
Thus, substituting the values, we have:
c (x) = (5.5x + k) / (x)
9 = (5.5 (200) + k) / (200)
Clearing K
(200) * 9 = 5.5 (200) + k
k = (200) * 9 - 5.5 * (200)
k = 700
answer
the value of the constant k is 700

3 0

3 years ago

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