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3 years ago

Simplify 5 – 2x – 3 + x.

Mathematics

1 answer:

g100num [7]3 years ago

7 0

Hey there!

All we have to do is combine like terms. Therefore, your final answer would be 2 - x

Hope this helps you!

God bless ❤️

xXxGolferGirlxXx

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(t-distribution) A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. To m

statuscvo [17]

Answer:

There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. This means that $\mu = 50$ .

We found that the sample had a average lifespan of 47.3 months, and a standard deviation of s = 9 months. What is the probability that we find this lifespan for our sample average, or something even shorter?

We have to find the pvalue of Z when $X = 47.3$ .

We are working with a sample mean, so we use the standard deviation of the sample in the place of $\sigma$ . That is $s = 9$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{47.3-50}{9}$

$Z = -0.3$

$Z = -0.3$ has a pvalue of 0.3821.

There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.

7 0

3 years ago

HELPP!!! Isn't one of the answers supposed to be 3?????

ikadub [295]

Step-by-step explanation:

well, it is supposed to be -3.

but you are right. something is wrong or missing here.

5x + 3 = 4x

x + 3 = 0

x = -3

maybe a "+ 1" is missing on the right side ?

5x + 3 = 4x + 1

gives us

x = -2

7 0

2 years ago

Read 2 more answers

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%