1.The data set shows the October 1 noon temperatures in degrees Fahrenheit for a particular city in each of the past 6 years.

The triangle below are drawn on 1-cm dot paper. find the perimeter of each triangle

AnnyKZ [126]

There is no triangle below. Try to edit this problem.

4 0

3 years ago

Evaluate the expression below when y = -3. 2y + 7 -4y − 10

sveticcg [70]

The problem asks:

what is 2y+7 when y=-3?

Since y=-3 then 2y+7=2(-3)+7=-6+7=1.

Similarly, when y=-3, -4y − 10=-4(-3) − 10=12-10=2.

Answer:

2y + 7=1

-4y − 10=2

7 0

3 years ago

Read 2 more answers

17) 10(x + 3) -(-9x - 4) = x-5 +3

navik [9.2K]

Answer:

-2

Step-by-step explanation:

10x+30+9x+4=x-5+3

19x+34=x-2

18x=-36

x=-2

5 0

4 years ago

The sum of two numbers is 54. The large number is 18 more than the sampler number. What are the numbers ?

Karolina [17]

Answer: a=36; b=18

Step-by-step explanation:

1. You know that the sum of two numbers is 54, then, you have:

a+b=54

2. According to the problem, the larger number is 18 more than the sampler number, this can be expressed as following:

a=b+18

3. Then, you must substitute the second equation into the first equation and solve for b:

b+18+b=54

b=18

4. Then the value of a is:

a+18=55

a=36

7 0

4 years ago

Construct a confidence interval that will capture the true population mean with 95% confidence. Population standard deviation is

kompoz [17]

A confidence interval that will capture the true population mean with 95% confidence. The value is ( 23.6459 , 24.8293 ) .

What is population mean?

A population in statistics is a group of comparable objects or occurrences that are relevant to a particular topic or experiment. A statistical population can be a collection of real things or a hypothetical, possibly limitless collection of objects derived from experience.

Data: 18

for given data

$\begin{aligned}\bar{X} & =\frac{\sum X i}{n} \\& =2448 / 101 \\& =24.2376\end{aligned}$

as population variance is unknown so we used sample variance

$\begin{aligned}S^2 & =\frac{1}{n-1}\left(\sum X i^2-n \times \bar{X}^2\right) \\S^2 & =\frac{1}{100}\left(60232-101 \times 24.2376^2\right) \\& =8.9830\end{aligned}$

Now

$\frac{\bar{X}-\mu}{S / \sqrt{n}} \sim t_{n-1}$$$

hence 95 % of population mean is

As n=101 which is large so critical value is 1.98397

$\begin{aligned}& =\left(\bar{X}-\sqrt{\frac{S}{n}} \times t_{(n-1, \alpha / 2)}, \bar{X}+\sqrt{\frac{S}{n}} \times t_{(n-1, \alpha / 2)}\right) \\& =\left(24.2376-\sqrt{\frac{8.9830}{101}} \times 1.98397,24.2376+\sqrt{\frac{8.9830}{101}} \times 1.98397\right) \\& =(23.6459,24.8293)\end{aligned}$

To learn more about statistics visit:brainly.com/question/29093686

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4 0

1 year ago