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ra1l [238]
3 years ago
14

Giving brainliest to whoever answers this.

Mathematics
1 answer:
hjlf3 years ago
8 0

Answer:

5,1

Step-by-step explanation:

7x + 4y = 39

2x + 4y = 14

since, the 4y same for both equation then we can eliminate the 4y by subtracting the two equation

7x + 4y = 39

2x + 4y = 14

---------------- -

5x + 0 = 25

5x = 25

x = 25/5 = 5

then we can input (subtitute) the value of x into the equation, I choose 2x + 4y = 14

2x + 4y = 14

2(5) + 4y = 14

10 + 4y = 14

4y = 14 - 10

4y = 4

y = 4/4 = 1

or if your teacher ask you to use elimination to find the value of x we can write like this

7x + 4y = 39

2x + 4y = 14

make the coeficient of x in both equation same

7x + 4y = 39 (x2)

2x + 4y = 14 (x7)

so

14x + 8y = 78

14x + 28y = 98

--------------------- -

0 - 20y = -20

-20y = -20

y = (-20)/(-20) = 1 (same answer right)

so the correct answe is

(5,1)

You might be interested in
Each year about 1500 students take the introductory statistics course at a large university. This year scores on the nal exam ar
nikitadnepr [17]

Answer:

a) Left-skewed

b) We should expect most students to have scored above 70.

c) The scores are skewed, so we cannot calculate any probability for a single student.

d) 0.08% probability that the average score for a random sample of 40 students is above 75

e) If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.

Step-by-step explanation:

To solve this question, we need to understand skewness,the normal probability distribution and the central limit theorem.

Skewness:

To undertand skewness, it is important to understand the concept of the median.

The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.

If the median is larger than the mean, the distribution is left-skewed.

If the mean is larger than the median, the distribution is right skewed.

Normal probabilty distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation, also called standard error of the mean s = \frac{\sigma}{\sqrt{n}}

(a) Is the distribution of scores on this nal exam symmetric, right skewed, or left skewed?

Mean = 70, median = 74. So the distribution is left-skewed.

(b) Would you expect most students to have scored above or below 70 points?

70 is below the median, which is 74.

50% score above the median, and 50% below. So 50% score above 74.

This means that we should expect most students to have scored above 70.

(c) Can we calculate the probability that a randomly chosen student scored above 75 using the normal distribution?

The scores are skewed, so we cannot calculate any probability for a single student.

(d) What is the probability that the average score for a random sample of 40 students is above 75?

Now we can apply the central limit theorem.

\mu = 70, \sigma = 10, n = 40, s = \frac{10}{\sqrt{40}} = 1.58

This probability is 1 subtracted by the pvalue of Z when X = 75. So

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

By the Central limit theorem

Z = \frac{X - \mu}{s}

Z = \frac{75 - 70}{1.58}

Z = 3.16

Z = 3.16 has a pvalue of 0.9992

1 - 0.9992 = 0.0008

0.08% probability that the average score for a random sample of 40 students is above 75

(e) How would cutting the sample size in half aect the standard error of the mean?

n = 40

s =  \frac{10}{\sqrt{40}} = 1.58

n = 20

s =  \frac{10}{\sqrt{20}} = 2.24

If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.

4 0
3 years ago
Please answer .....
melisa1 [442]

Answer:

A) 5/12

B) 5/12

C) 0/12

D) 11/12

-------------

Q4.

A) 3/15

B) 4/15

C) 12/15

D) 8/15

-----------------

Q5.

A) 2/6 or 1/3

B) 4/6 or 2/3

C)

D)

Step-by-step explanation:

6 0
2 years ago
(1 point) A recent report for a regional airline reported that the mean number of hours of flying time for its pilots is 56 hour
Ray Of Light [21]

Answer:

(53.812 ; 58.188) ; 156

Step-by-step explanation:

Given that :

Sample size (n) = 51

Mean (m) = 56

Standard deviation (σ) = 9.5

α = 90%

Using the relation :

Confidence interval = mean ± Error

Error = Zcritical * (standard deviation / sqrt (n))

Zcritical at 90% = 1.645

Error = 1.645 * (9.5 / sqrt(51))

Error = 1.645 * 1.3302660

Error = 2.1882877

Hence,

Confidence interval :

Lower boundary = 56 - 2.1882877 = 53.8117123

Upper boundary = 56 + 2.1882877 = 58.1882877

Confidence interval = (53.812 ; 58.188)

2.)

Margin of Error (ME) = 1.25

α = 90%

Sample size = ((Zcritical * σ) / ME)^2

Zcritical at 90% = 1.645

Sample size = ((1.645 * 9.5) / 1.25)^2

Sample size = (15.6275 / 1.25)^2

Sample size = 12.502^2 = 156.3000

Sample size = 156

3 0
3 years ago
Solve the following equation for a. be sure to take into account whether a letter is capitalized or not g = Ha + na
dimulka [17.4K]

Answer:

a = \frac{g}{H+n}

Step-by-step explanation:

Step 1: Factor

g = a(H + n)

Step 2: Divide both sides by expression in parenthesis

g/(H + n) = a

7 0
4 years ago
Plz help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Harrizon [31]
You at the photos and what you see put it down and what you know put it down
8 0
3 years ago
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