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zheka24 [161]
3 years ago
12

Please help me asap​

Mathematics
1 answer:
Valentin [98]3 years ago
7 0

Answer:

third option

Step-by-step explanation:

(five plus)= 5+

(the quotient of r and 14)=r/14

i think... :)

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The perimeter of a triangle is 7 m. The longest side is twice the length of the shortest side. The sum of the
34kurt

Answer:

See below in bold.

Step-by-step explanation:

Let the length of the shortest side be x, then the longest side = 2x.

Also  x + y = 2x  + 1   where y = length of the middle side,

Simplifying:

2x - x - y = -1

x - y = -1        (A)

Also as the perimeter = 7:

x + 2x + y = 7

3x + y = 7       (B)

Adding (A) and (B):

4x + 0 = 6

x = 6/4 = 1.5

Plug x = 1.5 into equation B:

3(1.5) + y = 7

y = 7 - 4.5 = 2.5.

Answer:

So the shortest side = 1.5, middle = 2.5 and longest = 2*1.5 = 3.

3 0
2 years ago
Help please!(picture)
UkoKoshka [18]

Answer:

7a - 2

Step-by-step explanation:

A perimeter is just the sum of all sides added together, so just put all the expressions together into a big equation.

2a - 3 + 2a + 3a + 1

Combine like terms

7a - 2

5 0
3 years ago
The number of customers that come to a certain clothing store each day follows a normal distribution. The mean number of custome
Lera25 [3.4K]

Answer:

c 2.5

Step-by-step explanation:

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3 years ago
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-10-(-6) solve the integers
Ghella [55]

Answer:

It would be -4

Step-by-step explanation:

a double negative becomes a + sign

So, -10 + 6=-4

4 0
3 years ago
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A study of 25 graduates of 4-year public colleges revealed the mean amount owed by a student in student loans was $55,051. The s
Harman [31]

Answer:

Step 1

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

Step 2

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

5 0
2 years ago
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