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blondinia [14]
3 years ago
10

The standard deviation of the lifetime of a particular brand of car battery is 4.3 months. A random sample of 10 of these batter

ies is taken. Assuming that the lifetimes of all batteries of this brand are normally distributed, what is the probability that the mean lifetime for this sample is within 1 mouth of the mean lifetime for all batteries of this brand?
a. 0.019
b. 0.184
c. 0.519
d. 0.538
e. 0.769
Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

d. 0.538

Step-by-step explanation:

Given information.

Standard deviation is 4.3

Random sample of 10

Consider the following calculations

P (sample mean is within 1 month of population mean)=

P(-1*sqrt(10)/4.3 <Z<sqrt(10)/4.3)=P(-.735<Z<0.735)

=0.538

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Stephanie has $152 in the bank she with drawl $20 then she deposits $84 write an addition expression to represent this situation
Paraphin [41]

152+(-20)+84=

236+(-20)=

236-20=

216

3 0
3 years ago
How much does a glass cube with a side of 5-cm weigh if 1cm3 of glass weighs 2 2/5 grams?
PtichkaEL [24]

The weight of galss cube with side 5 cm is 300 grams

<em><u>Solution:</u></em>

A glass cube is of side length 5 cm

Let us first find the volume of glass cube

<em><u>The volume of cube is given as:</u></em>

volume = (side)^3\\\\volume =5^3 = 125

Thus volume of cube is 125 cm^3

Given that,

1 cm^3 = 2\frac{2}{5} \text{ grams }\\\\1 cm^3 =\frac{12}{5} \text{ grams}

So, weight of 125 cubic centimeter of glass cube is found by multiplying weight of 1 cubic centimeter of glass cube by 125

\text{ Weight of 125 } cm^3 \text{ cube } = \text{ Weight of 1 } cm^3 \text{ cube } \times 125

\text{ Weight of 125 } cm^3 \text{ cube } = \frac{12}{5} \times 125 = 25 \times 12 = 300

Thus weight of galss cube with side 5 cm is 300 grams

3 0
3 years ago
Read 2 more answers
Solve x2+12x=-11 by completing the square . Which is the the solution set of the equation?
Dmitriy789 [7]

Answer:

{- 11, - 1 }

Step-by-step explanation:

Since the coefficient of the x² term is 1 , to complete the square

add (half the coefficient of the x-term )² to both sides

x² + 2(6)x + 36 = - 11 + 36 ← complete the square on the left side

(x + 6)² = 25 ( take the square root of both sides

\sqrt{(x+6)^2} = ± \sqrt{25}

x + 6 = ± 5 ( subtract 6 from both sides )

x = - 6 ± 5

x = - 6 + 5 = - 1 or x = - 6 - 5 = - 11




8 0
3 years ago
Read 2 more answers
What is the greatest common factor of 11,44?
mylen [45]
11, because 11 times 1 is 11 and 11 times 4 is 44.
4 0
3 years ago
The average expenditure on Valentine's Day was expected to be$100.89 (USA Today, February 13, 2006). Do male and femaleconsumers
stealth61 [152]

Answer:

(a) $62.16

(b) Male: $15.00

Female: $10.06

(c) Confidence Interval for male expenditure is ($106.40, $136.40)

Confidence interval for female expenditure is ($49.18, $69.30)

Step-by-step explanation:

(a) Male expenditure

Sample mean = $135.67, sd=$35, n=40, Z=2.576

Population mean = sample mean - (Z×sd)/√n = 135.67 - (2.576×35)/√40 = 135.67 - 14.27 = $121.40

Female expenditure

Sample mean= $68.64, sd=$20, n=30, Z=2.576

Population mean = 68.64 - (2.576×20)/√30 = 68.64 - 9.40 = $59.24

$121.40 - $59.24 = $62.16

(b) Male: Error margin = (t-value × sd)/√n

Degree of freedom = n-1 = 40-1= 39. t-value corresponding to 39 degrees of freedom and 99% confidence level is 2.708

Error margin = (2.708×35)/√40 = 94.78/6.32 = $15.00

Female

Degrees of freedom = n-1 = 30-1 = 29. t-value is 2.756

Error margin = (2.756×20)/√30 = 55.12/5.48 = $10.06

(c) Male

Confidence Interval (CI) = (mean + or - error margin)

CI = 121.4 + 15.00 = $136.40

CI = 121.4 - 15.00 = $106.40

Confidence Interval is ($106.40, $136.40)

Female

CI = 59.24 + 10.06 = $69.30

CI = 59.24 - 10.06 = $49.18

Confidence Interval is ($49.18, $69.30)

7 0
3 years ago
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