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4 years ago

Is it possible to multiply a 3 by 4 matrix and a 2 by 2 matrix

1 answer:

8 0

No, because the number of rows of the first one has to equal the number of collumns in the 2nd matrix

none of them are equal (except 2=2 but same matrix0

cannot multiply

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Thee population of a town, P(t), is 2000 in the year 2010. Every year, it increases by 1.5%. Write an equation for the populatio

steposvetlana [31]

Given:

In 2010, the population of a town = 2000

Every year, it increase by 1.5%

To find the equation P(t) which represents the population of this town t years from 2010.

Formula

If a be the original population of a town and it increases by b% every year, after t year the population will be

$a (1+\frac{b}{100} )^{t}$

Now,

Taking, a =2000, b = 1.5 we get,

$P(t) = 2000(1+\frac{1.5}{100} )^{t}$

or, $P(t) = 2000(1+0.015)^{t}$

Hence,

The population of this town t years from 2010 P(t) = $2000(1+0.015)^{t}$ , Option D.

7 0

3 years ago

(c^3b^1a^4)^4 + 3a^16b^4c^12

azamat

Answer: 4a16b4c12

Step-by-step explanation:

5 0

3 years ago

1) What can u do to get 2x alone on the left side?

CaHeK987 [17]

Answer:

82.5 - 82.5 + 2x = 338.5 - 82.5

2x = 256

2x ÷ 2 = 256 ÷ 2

x = 128

Step-by-step explanation:

In order to get rid 82.5, you have to substract the same value on both side to get a 0 value.

In order to find the value of x, you have to divide the value that is sticked with x which is 2. As you divide 2 by 2, you will get 1 as a single value of x so you have to divide 2 to both side too.

5 0

3 years ago

GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m

Harrizon [31]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

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

where, $\bar X$ = sample mean mg of mercury = 3.59

$\sigma$ = population standard deviation = 0.18 mg

n = sample of bulbs = 25

So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

= 2.50

The value of z test statistics is 2.50.

Now, P-value of the test statistics is given by;

P-value = P(Z > 2.50) = 1 - P(Z $\leq$ 2.50)

= 1 - 0.9938 = 0.0062

Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0

3 years ago

Write the smallest and the largest 5 - digit number using 1 ,2,3,4,5 only one which is divisble by 6

Aleks04 [339]

Largest 5-digit number is 54312 and is divisible by 6. The smallest 5 digit number is 12,345 and it's not divisible by 6

4 0

3 years ago