-0.5y-2(-5y+9) please help! i will give you 12 points

Sarah baby-sat 7 times as many hours during summer break as she did during spring break. She baby-sat a total of 56 hours during

kotykmax [81]

I wanna say the answer could be 8

3 0

4 years ago

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A(n) = -5 + 6(n - 1)

Andreas93 [3]

Answer:

domain is all reall numbers

3 0

3 years ago

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The pet doctor weighs Marc's dog every year. Last year, the dog lost 3.73 pounds from the previous year. This year, the dog gain

levacccp [35]

Answer:

-2.26 lb

Step-by-step explanation:

Let the dog's initial weight be d.

Then, one year later, the dog weighs 3.73 lb less, and thus weighed d - 3.73 lb.

During the next year, the dog gained 1.47 lb, so now weighs d - 3.73 + 1.47 lb, or

d - 2.26 lb. This indicates a total weight loss of 2.26 over these two years.

6 0

3 years ago

67, 567 x 47 a 317, 649 b 3,175,649 c 2,274, 549 d 3,863, 67

Anit [1.1K]

Answer:

option a 317,649

Step-by-step explanation:

please follow me

4 0

3 years ago

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Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply,

myrzilka [38]

Answer:

$163.83-4.03\frac{20.094}{\sqrt{6}}=130.77$

$163.83+4.03\frac{20.094}{\sqrt{6}}=196.89$

So on this case the 99% confidence interval would be given by (130.77;196.89)

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 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".

Data: 175 177 175 180 138 138

We can calculate the mean and the deviation from these data with the following formulas:

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

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

$\bar X=163.83$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=20.093 represent the sample standard deviation

n=6 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=6-1=5$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,5)".And we see that $t_{\alpha/2}=4.03$

Now we have everything in order to replace into formula (1):

$163.83-4.03\frac{20.094}{\sqrt{6}}=130.77$

$163.83+4.03\frac{20.094}{\sqrt{6}}=196.89$

So on this case the 99% confidence interval would be given by (130.77;196.89)

7 0

3 years ago