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2 years ago

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For a week Raymond is paid at the rate of x dollars per hour for the first 40 hours he works and y dollars per hour for the hour

s worked in excess of 40 hours. What is the value of y?

1 answer:

Andre45 [30]2 years ago

8 0

Raymond would be paid $Y/h whenever he works over 40 hours.

Let's say the number of hours he works is "h."

Assuming H>40,
Y(H-40)

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The amount of profit, p, you earn by selling knives, k, can be determined by: p=200k-500 a) Determine the constraints on profit

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We are given with the formula
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If ∆ABC and ∆GEF are congruent, the value of x is___? and the value of y is ___?

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<u>Answer-</u>

<em>The value of y is </em><em>6 units</em><em>.</em>

<u>Solution-</u>

Given ∆ABC ≅ ∆GEF

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EF = BC, GF = AC

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2 years ago

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A random sample from a normal population is obtained, and the data are given below. Find a 90% confidence interval for . 114 157

melamori03 [73]

Answer:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

Step-by-step explanation:

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 mean or variance 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.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

Data given: 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545

The confidence interval for the population variance $\sigma^2$ is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^{17} (x_i -\bar x)^2}{n-1}}$

And in order to find the sample mean we just need to use this formula:

$\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample mean obtained on this case is $\bar x= 362.941$ and the deviation s=132.250

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=17-1=16$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a tabel to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(0.95,16)". so for this case the critical values are:

$\chi^2_{\alpha/2}=26.296$

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And replacing into the formula for the interval we got:

$\frac{(16)(132.250)^2}{26.296} \leq \sigma \leq \frac{(16)(132.250)^2}{7.962}$

$10641.959 \leq \sigma^2 \leq 35147.074$

Now we just take square root on both sides of the interval and we got:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

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Julli [10]

For this case we have the following equation:

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From here, we clear the value of x.

We have then:

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Then, we look for the value of 2x. We have then:

$2x = 2 \frac {1} {2}\\2x = \frac {2} {2}\\2x = 1$

Answer:

The value of 2x is given by:

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