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3 years ago

6

After each charging, a battery is able to hold only 99% of the charge from the previous charging. The battery was used for 15 ho

urs on its first charge before it had to be recharged. What is the total number of hours the battery can be used over its lifetime?

1 answer:

valentina_108 [34]3 years ago

4 0

Answer: 13.5 hours or 14 hours (if you round it)

Step-by-step explanation:

$\frac{0.90}{1.00}$ * 15

=13.5

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The fraction is throwing me off..help..!

svetlana [45]

Answer:

x=-21

Step-by-step explanation:

1/3 x + 9 =2

Subtract 9 from each side

1/3 x +9-9 =2-9

1/3x = -7

Multiply by 3 to isolate x

3*1/3x = -7*3

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4 years ago

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3 years ago

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The power is the probability that the assembly process is not stopped when it should be because a high number of defective items

choli [55]

Answer:

B. The power is the probability that the assembly process continues because the proportion of items that must be rejected has not increased.

Step-by-step explanation:

The probability is the percentage which determines the occurrence of an event. The probability measure is of power determines that the assembly process will continue because the proportion of the items that must be rejected is not increased.

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Two different types of polishing solutions are being evaluated for possible use in a tumble-polish operation for manufacturing i

Nina [5.8K]

Answer:

$z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358$

$p_v =2*P(Z>5.358) = 4.2x10^{-8}$

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.

Step-by-step explanation:

Data given and notation

$X_{1}=253$ represent the number with no defects in sample 1

$X_{2}=196$ represent the number with no defects in sample 1

$n_{1}=300$ sample 1

$n_{2}=300$ sample 2

$p_{1}=\frac{253}{300}=0.843$ represent the proportion of number with no defects in sample 1

$p_{2}=\frac{196}{300}=0.653$ represent the proportion of number with no defects in sample 2

z would represent the statistic (variable of interest)

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

$\alpha=0.05$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference in the the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} - p_2}=0$

Alternative hypothesis: $p_{1} - p_{2} \neq 0$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{253+196}{300+300}=0.748$

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358$

Statistical decision

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

$p_v =2*P(Z>5.358) = 4.2x10^{-8}$

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.

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4 years ago