All categories

3 years ago

Can a matrix of dimensions 2 X 4 be added to another matrix with dimensions of 2 X 6?

Mathematics

1 answer:

Rus_ich [418]3 years ago

8 0

Answer:

No.

Step-by-step explanation:

It has different order of matrices .

For Addition or Substraction , both matrices must have the same number of rows and columns .

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Find the quadratic function y=a(x-h)^2 whose graph passes through the given points. (12,-7) and (9,0)

ryzh [129]

Answer:

y = (-7/3)(x - 9)

Step-by-step explanation:

(12, - 7)

y = a(x - h)^2

-7 = a(12 - h)^2

- 7 = a(144 - 24x + h^2)

(9,0)

0 = a(x - h)^2

0 = a(9 - h)^2

0 = a(81 - 18h + h^2)

From (9,0) we can conclude that

a = 0

(9 - h)^2 = 0

Let's try the second possibility.

Take the square root of both sides.

9 - h = sqrt(0)

Add h to both sides

9 - h = 0

h = 9

So now what we have is

y = a(x - 9)

Use the first equation to get a

-7 = a(12 - 9)

-7 = a(3)

-7/3 = a

Answer

y = (-7/3)(x - 9)

6 0

3 years ago

A store charges a restocking fee for any returned item based upon the item price. An item priced at $200 has a fee of $12. An it

lyudmila [28]

Answer:

Both ratios reduce to the same ratio 3/50, so the restocking fee is proportional.

Step-by-step explanation:

For the $200, the restocking fee is $12, so the ratio of the restocking fee to the price of the item is 12/200.

For the $150, the restocking fee is $9, so the ratio of the restocking fee to the price of the item is 9/150.

Now we find out if the ratios 12/200 and 9/150 are equal.

12/200 = 3/50

9/150 = 3/50

Both ratios reduce to the same ratio 3/50, so the restocking fee is proportional.

7 0

2 years ago

PLEASE I NEWD HELP ON THIS

Zarrin [17]

Answer:

5. Inequality form: n ≥ 3

Interval notation: [3, ∞)

6. Inequality form: x < 4

Interval notation: (- ∞, 4)

11. Inequality form: x > 50

Interval notation: (50, ∞)

12. Inequality form: y $\geq$ 8

Interval notation: [8, ∞)

15. Inequality form: z < 8

Interval notation: (- ∞, 8)

16. Inequality form: y $\geq$ 4

Interval notation: [4, ∞)

Step-by-step explanation:

I tried to graph them as best as I could, I hope this helps!

7 0

3 years ago

A college requires applicants to have an ACT score in the top 12% of all test scores. The ACT scores are normally distributed, w

DochEvi [55]

Answer:

a) The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b) 156 would be expected to have a test score that would meet the colleges requirement

c) The lowest score that would meet the colleges requirement would be decreased to 26.388.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 21.5, \sigma = 4.7$

a. Find the lowest test score that a student could get and still meet the colleges requirement.

This is the value of X when Z has a pvalue of 1 - 0.12 = 0.88. So it is X when Z = 1.175.

$Z = \frac{X - \mu}{\sigma}$

$1.175 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.175*4.7$

$X = 27.0225$

The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b. If 1300 students are randomly selected, how many would be expected to have a test score that would meet the colleges requirement?

Top 12%, so 12% of them.

0.12*1300 = 156

156 would be expected to have a test score that would meet the colleges requirement

c. How does the answer to part (a) change if the college decided to accept the top 15% of all test scores?

It would decrease to the value of X when Z has a pvalue of 1-0.15 = 0.85. So X when Z = 1.04.

$Z = \frac{X - \mu}{\sigma}$

$1.04 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.04*4.7$

$X = 26.388$

The lowest score that would meet the colleges requirement would be decreased to 26.388.

6 0

4 years ago

Can someone help me with question 9

Troyanec [42]

Please refer the pictures below-

Hence the area of the road is 9600m2

8 0

3 years ago