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2 years ago

BRAINLY HELP NEEDED ASAP

Mathematics

1 answer:

forsale [732]2 years ago

7 0

Step-by-step explanation:

increase in Tyler's pay = 16 - 14 = 2

% increase in Tyler's pay = 2/14 × 100 = 14.3%

Increase in Allies' pay = 18 - 16 = 2

% increase in Allies'pay = 2/16 × 100 = 12.5%

hence the right option is:

3).The percent increase Tyler's pay rate is greater than Allies'.

Increase in Martin's pay = 15.00 - 13.50 = 1.50

% increase in Martin's pay = 1.50/13.50 = 11.1%

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Suppose the scores of students on an exam are normally distributed with a mean of 244 and a standard deviation of 79. According

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244 - 165 = 79 . . . . one standard deviation
323 - 244 = 79 . . . . one standard deviation

The range of scores is ±1 standard deviation from the mean. The empirical rule says
68% of scores lie in that range.

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3 years ago

What is 5(2 + x) = 5 ?

Alex787 [66]

5(2+x)=5

or, 10+5x=5

or, 5x= -5

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3 years ago

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A punch recipe requires 2 cups of cranberry juice to make 3 gallons of punch. Using the same recipe, what is the amount of cranb

gogolik [260]

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3 years ago

If there was a total of 30 computers and the computer lab is removing 3 broken computers and adding 13 wireless devices, about w

Liula [17]

32.5 % of lab are wireless devices

Solution:

Given that there was a total of 30 computers and the computer lab is removing 3 broken computers and adding 13 wireless devices

Now we can first find out total number of computers and devices in lab

total number of computers and devices in lab = 30 computers - 3 + 13 wireless devices

total number of computers and devices in lab = 30 - 3 + 13 = 40

So there are 40 computers and wireless devices in lab

To find: what percent of the lab are wireless devices

So we have to find what percent of 40 is 13

Let "a" be the required percent

a % of 40 = 13

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32.5 % of of lab are wireless devices

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3 years ago

Granola Crunch cereal is packaged in 1 pound boxes. Susan Torres, a quality control analyst who works for the manufacturer of th

sineoko [7]

Answer:

$z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162$

$p_v =2*P(z>3.162)=0.0016$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=1.02$ represent the sample mean

$\sigma=0.04$ represent the population standard deviation

$n=40$ sample size

$\mu_o =1$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is equal to 1 pound or no :

Null hypothesis: $\mu =1$

Alternative hypothesis: $\mu \neq 1$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162$

P-value

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

$p_v =2*P(z>3.162)=0.0016$

Conclusion

5 0

3 years ago

**BRAINLY** HELP NEEDED ASAP

BRAINLY HELP NEEDED ASAP