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chubhunter [2.5K]
3 years ago
12

A plumber charges $55.00 for a service call plus $35.00 per hour. If the total bill was $160.00 . Select an equation , with x re

presenting the number of hours that can be use to determine how many hours the plumber was there.
A) 55x + 35 = 160
B) 35x + 160 = 55
C) 35x + 55 = 160
D) 160x +55 = 35.00
Mathematics
2 answers:
Vladimir [108]3 years ago
7 0

Answer:

its C that's definitely correct

Anestetic [448]3 years ago
7 0
Pretty sure its C if not I’m very sorry
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10<br> 9. Solve for x.<br> (9x + 1)<br> 88<br> PLEASE HELP! will mark brainliest
Varvara68 [4.7K]

Answer:

x = 9.6

Step-by-step explanation:

9x + 1 = 88

combine like terms

9x = 87

divide

87/9 = 9.6

x = 9.6

3 0
3 years ago
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Given C=n+15, determine the value of C when n=4
puteri [66]

Answer: c=19 and n=4

please mark as brainliest  if it helped

6 0
3 years ago
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Read this E[2X^2 â€" Y].
djyliett [7]

Looks like a badly encoded/decoded symbol. It's supposed to be a minus sign, so you're asked to find the expectation of 2<em>X </em>² - <em>Y</em>.

If you don't know how <em>X</em> or <em>Y</em> are distributed, but you know E[<em>X</em> ²] and E[<em>Y</em>], then it's as simple as distributing the expectation over the sum:

E[2<em>X </em>² - <em>Y</em>] = 2 E[<em>X </em>²] - E[<em>Y</em>]

Or, if you're given the expectation and variance of <em>X</em>, you have

Var[<em>X</em>] = E[<em>X</em> ²] - E[<em>X</em>]²

→   E[2<em>X </em>² - <em>Y</em>] = 2 (Var[<em>X</em>] + E[<em>X</em>]²) - E[<em>Y</em>]

Otherwise, you may be given the density function, or joint density, in which case you can determine the expectations by computing an integral or sum.

6 0
3 years ago
50 people are selected randomly from a certain population and it is found that 12 people in the sample are over 6 feet tall. wha
Taya2010 [7]

Answer:

The point estimate of the proportion of people over 6ft to total people is 6:25

Step-by-step explanation:

To find this, simply start with the number of people over 6 ft and total people as a proportion.

12:50

Then simplify by dividing by 2

6:25

8 0
3 years ago
The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance
lisabon 2012 [21]

Answer:

The 95% confidence interval for the variance in the pounds of impurities would be 3.829 \leq \sigma^2 \leq 14.121.

Step-by-step explanation:

1) Data given and notation

s^2 =6.62 represent the sample variance

s=2.573 represent the sample standard deviation

\bar x represent the sample mean

n=20 the sample size

Confidence=95% or 0.95

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 .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

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

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

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

df=n-1=20-1=19

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:

\chi^2_{\alpha/2}=32.852

\chi^2_{1- \alpha/2}=8.907

And replacing into the formula for the interval we got:

\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}

3.829 \leq \sigma^2 \leq 14.121

So the 95% confidence interval for the variance in the pounds of impurities would be 3.829 \leq \sigma^2 \leq 14.121.

4 0
4 years ago
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