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3 years ago

The sum of two integers is -7. The product of the integers is -18. What are the integers

1 answer:

6 0

Let's solve this problem step-by-step.

STEP-BY-STEP SOLUTION:

We will be using simultaneous equations to solve this problem.

Let's first establish the two equations we will be using.

Let integer no. 1 = x

Let integer no. 2 = y

Equation No. 1 -

x + y = - 7

Equation No. 2 -

xy = - 18

To begin with, we will make ( x ) the subject in the first equation.

Equation No. 1 -

x + y = - 7

x = - 7 - y

Next we will substitute the value of ( x ) from the first equation into the second equation to solve for ( y ).

Equation No. 2 -

xy = - 18

y ( - 7 - y ) = - 18

- 7y - y^2 = - 18

- y^2 - 7y + 18 = 0

- [ y^2 + 7y - 18 ] = 0

- [ y^2 + 9y - 2y - 18 ] = 0

- [ y ( y + 9 ) - 2 ( y + 9 ) ] = 0

- ( y + 9 ) ( y - 2 ) = 0

y + 9 = 0

y = - 9

OR

y - 2 = 0

y = 2

Then we will substitute the value of ( y ) from the second equation into the first equation to solve for ( x ).

Equation No. 1 -

x = - 7 - y

x = - 7 - ( - 9 )

x = - 7 + 9

x = 2

OR

x = - 7 - y

x = - 7 - ( 2 )

x = - 9

FINAL ANSWER:

Therefore, the answer is:

x = 2

y = - 9

OR

x = - 9

y = 2

The two integers are 2 and - 9 or - 9 and 2.

Please mark as brainliest if you found this helpful! :)
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3 years ago

In a certain year, when she was a high school senior, Idonna scored 671 on the mathematics part of the SAT. The distribution of

goldfiish [28.3K]

Answer:

Idonna's standardized score is 1.41.

Jonathan's standardized score is 0.55.

A.) Idonna's score is higher than Jonathan's

Step-by-step explanation:

Z-score:

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.

Idonna scored 671 on the mathematics part of the SAT. The distribution of SAT math scores in that year was Normal with mean 509 and standard deviation 115.

This means that her standardized score is Z when $X = 671, \mu = 509, \sigma = 115$ . So

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

$Z = \frac{671 - 509}{115}$

$Z = 1.41$

Idonna's standardized score is 1.41.

Jonathan took the ACT and scored 24 on the mathematics portion. ACT math scores for the same year were Normally distributed with mean 21.1 and standard deviation 5.3 .

This means that his standardized score is Z when $X = 24, \mu = 21.1, \sigma = 5.3$

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

$Z = \frac{24 - 21.1}{5.3}$

$Z = 0.55$

Jonathan's standardized score is 0.55.

Due to the higher z-score, Iddona's has a higher score.

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3 years ago

What is the quotient

olasank [31]

Hope this would help you

8 0

3 years ago

Hi please help me with this i’ll give brainliest if you give a correct and show your work!

Tatiana [17]

1. (1.35)x(.42)

1.35
x. .42
________
270
5400
________
.5670

You need to have the total number of decimal places from the original two numbers. 1.35 has two decimal places and .42 has to decimal places you have a total of four decimal places, Which means 5670 is .5670

2. (.22)x(.04)

.22
x. .04
_______
88
00
_______
.0088

you need to have the total number of decimal places from the original Two numbers. So .22 is two decimal places and .04 is two decimal places, so you will need to have for decimal places. That means that the answer 88 needs to be translated into .0088

6 0

3 years ago

The average student-loan debt is reported to be $25,235. A student believes that the student-loan debt is higher in her area. Sh

yulyashka [42]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 25235

For the alternative hypothesis,

µ > 25235

This is a right tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100,

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 27524

µ = population mean = 25235

s = samples standard deviation = 6000

t = (27524 - 25235)/(6000/√100) = 3.815

We would determine the p value using the t test calculator. It becomes

p = 0.000119

Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.

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3 years ago