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Elden [556K]
1 year ago
7

Help Equation in the screenshot

Mathematics
1 answer:
Mama L [17]1 year ago
7 0

The value of point A21 in the matrix will be 6. The correct option is A.

<h3>What is a matrix?</h3>

A matrix, which is used to represent a mathematical object or an attribute of one, is a rectangular array or table containing numbers, symbols, or expressions that are organized in rows and columns.

The given matrix is:-

A=\left[\begin{array}{cccc}10&4&11&12\\6&3&2&8\\\end{array}\right]

The position of the numbers in the matrix is represented by the element Aij, where i is the number of the rows and j is the number of the columns.

For A₂₁ the position is the second row and the first column. At position A₂₁ the number is 6.

Therefore, the value of point A21 in the matrix will be 6. The correct option is A.

To know more about matrix follow

brainly.com/question/11989522

#SPJ1

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The price of an item increased from 20 to 30. The percentage of increase is pls help me I have exam after a week pls I hope you
Veronika [31]

Answer: 50% increase

Step-by-step explanation:

Think of 20 as 100%

Half of 20 is 10 which is 50% in this scenario.

30-20=10

This means that the price of the item increased by 50% or 10.

6 0
2 years ago
A sample of 11001100 computer chips revealed that 62b% of the chips fail in the first 10001000 hours of their use. The company's
STALIN [3.7K]

Answer:

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 90% confidence interval) ---- reject Null hypothesis

Z score < Z(at 90% confidence interval) ------ accept Null hypothesis

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

z score = 1.35

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

Question; A sample of 1100 computer chips revealed that 62% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 60% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.10 level.

Step-by-step explanation:

Given;

n=1100 represent the random sample taken

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 1100

po = Null hypothesized value = 0.60

p^ = Observed proportion = 0.62

Substituting the values we have

z = (0.62-0.60)/√{0.60(1-0.60)/1100}

z = 1.354

z = 1.35

To determine the p value (test statistic) at 0.01 significance level, using a two tailed hypothesis.

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

3 0
3 years ago
Can anyone help me with these three questions
iren2701 [21]
Sure, what are the questions?
5 0
3 years ago
According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones
Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = <u><em>back-to-college family spending on electronics</em></u>

SO, X ~ Normal(\mu=237,\sigma^{2} =54^{2})

The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean family spending = $237

           \sigma = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( \frac{X-\mu}{\sigma} < \frac{150-237}{54} ) = P(Z < -1.61) = 1 - P(Z \leq 1.61)

                                                             = 1 - 0.9463 = <u>0.0537</u>

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

        P(X > $390) = P( \frac{X-\mu}{\sigma} > \frac{390-237}{54} ) = P(Z > 2.83) = 1 - P(Z \leq 2.83)

                                                             = 1 - 0.9977 = <u>0.0023</u>

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X \leq $120)

     P(X < $175) = P( \frac{X-\mu}{\sigma} < \frac{175-237}{54} ) = P(Z < -1.15) = 1 - P(Z \leq 1.15)

                                                         = 1 - 0.8749 = 0.1251

     P(X < $120) = P( \frac{X-\mu}{\sigma} < \frac{120-237}{54} ) = P(Z < -2.17) = 1 - P(Z \leq 2.17)

                                                         = 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>

5 0
3 years ago
Jean spent $16 on new T-shirts. If each shirt cost the same whole-dollar amount, how many could she have bought?
Kaylis [27]
Well $16 times 16 would be $256.
If each shirt costs $1 then she could buy 16 shirts.
4 0
3 years ago
Read 2 more answers
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