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2 years ago

Write an equation for each problem. Use the variable M.

1 answer:

3 0

18) m=22+5. there would be 27 students.

19) k=59.95 - 22.05. kim pays 37.90

31) c = 3a. c=3(1.5). 4.5 pounds of cashews.

32) j=1/3m. j=1/3(45), or 45/3. josh is 15 years old.

You might be interested in

140 is 9% of what???

Len [333]

Answer:

The correct answer is B) 1555.6

Step-by-step explanation:

In order to find this, we must divide the part by the percentage. This will give us the total out of 100%

140/.09 = 1555.6

7 0

3 years ago

Determine if the columns of the matrix form a linearly independent set. Justify your answer. Choose the correct answer below. A.

kondaur [170]

Th matrix is missing. The matrix is :

$\begin{bmatrix}1 &-4 &4 \\ -4 &16 & 4 \end{bmatrix}$

Solution :

The column of the matrix are $\begin{bmatrix}1\\ -4\end{bmatrix}$ , $\begin{bmatrix}-4\\ 16\end{bmatrix}$ , $\begin{bmatrix}4\\ 4\end{bmatrix}$

Now each of them are vectors in $IR^2$ . But $IR^2$ has dimensions of 2. But there are 3 column vectors, hence they are linearly dependent.

Therefore, the column of the given matrix does not form the $\text{linearly independent set}$ as the set contains $\text{more vectors}$ than there are entries in each vector.

Therefore, option (D) is correct.

4 0

3 years ago

What would happen to the slope of the graph of y=2/3×-3 if the line were shifted 6 units up?

yaroslaw [1]

The slope wouldn't change, but the y-intercept would become 3 because you add -3 to 6

5 0

3 years ago

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago

3.52 in expanded form

Scrat [10]

Step-by-step explanation:

3.52 =

+ 0.5

+ 0.02

hopes this helps you

6 0

2 years ago