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Zolol [24]
2 years ago
13

Given the functions f(x) = 7x 13 and g(x) = x2 2, which of the following functions represents f[g(x)] correctly? f[g(x)] = 7x2 2

7 f[g(x)] = 7x2 15 f[g(x)] = 49x2 182x 169 f[g(x)] = 49x2 182x 171.
Mathematics
1 answer:
brilliants [131]2 years ago
7 0

Answer:

fg(×)=22fx

Step-by-step explanation:

I hope it help

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The corner store sells bananas at a rate of $2.50 per banana. If Jessica has $20.00 dollars, how many bananas can Jessica buy?
____ [38]

Answer:

8 bananas

Step-by-step explanation:

2.50b

20.00=2.50b

/2.50   /2.50

8=b

jessica can buy 8 bananas

4 0
3 years ago
Suppose Brianna decided that the error margin of her 95% confidence interval was too large and wanted an error margin of .05 whi
svetoff [14.1K]

Answer:

n=\frac{0.26 (1-0.26)}{(\frac{0.05}{1.96})^2} =295.65 \approx =296 (b)  

Rounded up would be n =296

Step-by-step explanation:

Assuming this previous info:  She randomly interested in the p selects 100 engineering majors, and 26 of them said they earned an A in Calc 1.

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

If solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)  

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:  

z_{\alpha/2}=\pm 1.96  

Thes estimated proportion is \hat p = \frac{26}{100}=0.26

And replacing we got:

n=\frac{0.26 (1-0.26)}{(\frac{0.05}{1.96})^2} =295.65 \approx =296 (b)  

3 0
3 years ago
Frank invested $30,000 at 4% simple interest. How much interest will he earn each year?
Oduvanchick [21]

Answer:

$1200 interest

Step-by-step explanation:

$30,000 divided by 100 = 300 , 300 multipled by 4 = 1200 , thats how to find what 4% would be

3 0
3 years ago
Type of car is considered to be
disa [49]

Answer:

non - living thing

Step-by-step explanation:

what the question

8 0
3 years ago
Read 2 more answers
Sick-leave time used by employees of a firm in a course of one month has approximately normal distribution, with a mean of 200 h
Usimov [2.4K]

Answer:

a)0.62% probability that total sick leave for next month will be less than 150 hours.

b) 225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

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:

\mu = 200, \sigma = \sqrt{400} = 20

a.Find the probability that total sick leave for next month will be less than 150 hours.

This probability is the pvalue of Z when X = 150. So:

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

Z = \frac{150 - 200}{20}

Z = -2.5

Z = -2.5 has a pvalue of 0.0062.

So there is a 0.62% probability that total sick leave for next month will be less than 150 hours.

b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

This is the value of X when Z has a pvalue of 0.90. So Z = 1.28

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

1.28 = \frac{X - 200}{20}

X - 200 = 20*1.28

X = 225.6

225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

6 0
4 years ago
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