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Degger [83]
3 years ago
12

John was told that his IQ score was 3 standard deviations above the mean. If IQ scores were approximately normal with μ=93 and σ

=9, what was John's score? Do not include units in your answer. For example, if you found that the score was 93 points, you would enter 93.
Mathematics
1 answer:
Darina [25.2K]3 years ago
4 0

9 is the standard deviation in this problem. The mean iq score is 93.

This meant that his IQ score is 93+9*3=93+27=120 :) also this john person is really smart im jealous :(

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15. Before oil is added to a certain road surface material, the other ingredients are mixed. The mixture contains
mojhsa [17]

Answer:

sand = 3450 pounds

3/4 inch stones =6750 pounds

1 inch stones =4800 pounds.

Step-by-step explanation:

The mixture composition is :

45% three-quarter inch stones

32% inch stones and

sand for balance

The percent here is by mass of material used.

Weight of mixture is 15,000 pounds before oil addition.

Use the percent by weight for the two types of stones and calculate real values in pounds

Formula to apply is:

Mass percent = (mass of a material/ total mass of mixture) * 100

For 3/4 inch stones it will be:

0.45= \frac{x}{15000} *100\\\\0.45=\frac{x}{150} \\0.45*150=x\\x=6750

The 3/4 inch stones weigh 6750 pounds

The 1 inch stones will be

32/100 * 15000 =4800

The 1 inch stones weigh = 4800 pounds

Total for the two items : 4800+6750 =11550 pounds

The sand used weighs : 15000 - 11550 = 3450 pounds

8 0
4 years ago
A baker puts 48 muffins Into boxes. Each box holds 5 muffins. To find the number of boxes needed, the baker completes the divisi
antoniya [11.8K]
There are 3 muffins that aren’t not in boxes
7 0
3 years ago
Factor y2 + 8y + 15
vovangra [49]
Y² + 8y + 15
since 8 and fifteen are positive, that means your factors are going to be positive
(y + )(y + )
What are the factors of 15 that add up to 8?
8 0
4 years ago
A2 = [1 2 3; 4 5 6; 7 8 9; 3 2 4; 6 5 4; 9 8 7]
34kurt

Answer:

Remember, a basis for the row space of a matrix A is the set of rows different of zero of the echelon form of A.

We need to find the echelon form of the matrix augmented matrix of the system A2x=b2

B=\left[\begin{array}{cccc}1&2&3&1\\4&5&6&1\\7&8&9&1\\3&2&4&1\\6&5&4&1\\9&8&7&1\end{array}\right]

We apply row operations:

1.

  • To row 2 we subtract row 1, 4 times.
  • To row 3 we subtract row 1, 7 times.
  • To row 4 we subtract row 1, 3 times.
  • To row 5 we subtract row 1, 6 times.
  • To row 6 we subtract row 1, 9 times.

We obtain the matrix

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&-6&-12&-6\\0&-4&-5&-2\\0&-7&-14&-5\\0&-10&-20&-8\end{array}\right]

2.

  • We subtract row two twice to row three of the previous matrix.
  • we subtract 4/3 from row two to row 4.
  • we subtract 7/3 from row two to row 5.
  • we subtract 10/3 from row two to row 6.

We obtain the matrix

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&0&0\\0&0&3&2\\0&0&0&2\\0&0&0&2\end{array}\right]

3.

we exchange rows three and four of the previous matrix and obtain the echelon form of the augmented matrix.

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&3&2\\0&0&0&0\\0&0&0&2\\0&0&0&2\end{array}\right]

Since the only nonzero rows of the augmented matrix of the coefficient matrix are the first three, then the set

\{\left[\begin{array}{c}1\\2\\3\end{array}\right],\left[\begin{array}{c}0\\-3\\-6\end{array}\right],\left[\begin{array}{c}0\\0\\3\end{array}\right] \}

is a basis for Row (A2)

Now, observe that the last two rows of the echelon form of the augmented matrix have the last coordinate different of zero. Then, the system is inconsistent. This means that the system has no solutions.

4 0
3 years ago
Giving brainliest!!!!!!!!!!
son4ous [18]
It’s A i did the test myself..
5 0
3 years ago
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