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Drupady [299]
3 years ago
5

One integer is 3 times another if the product of the two integers is 75 then find the integers

Mathematics
1 answer:
Kitty [74]3 years ago
6 0

Answer: HOPE THIS HELPED! :D

5 x 15 = 75

Step-by-step explanation:

15 is 3x 5

and product do 3x15 is 75

so answer would be

5    &     15

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A square has an area of 369.01m squared number.work out the perimeter of the square.
Lina20 [59]

Answer:

76.84 m is the perimeter

Step-by-step explanation:

First take the square root of the area

   √ 369.01 = 19.21  

Formula for  perimeter of square

         P = 4a

    a = 19.21

So  

        P = 4(19.21)

        P = 76.84 m

5 0
2 years ago
Factor each trinomial. Then match the polynomial (term) on the left with its factored form (definition) on the right.
sattari [20]
X^2 - 4x - 12 = x^2 + 2x - 6x - 12 = x(x + 2) - 6(x + 2) = (x - 6)(x + 2)
x^2 + 4x - 12 = x^2 - 2x + 6x - 12 = x(x - 2) + 6(x - 2) = (x - 2)(x + 6)
x^2 - x - 12 = x^2 + 3x - 4x - 12 = x(x + 3) - 4(x + 3) = (x - 4)(x + 3)
x^2 - 7x - 12 is prime.
4 0
3 years ago
What is the solution to the system of equations?<br><br> I will mark the brainliest answer
Stolb23 [73]
Ok so let us label our equations first
1- 3x+2y+z=7
2- 5x+5y+4z=3
3- 3x+2y+3z=1
subtracting equation 3 from equation 1
3x+2y+3z=1
(-). 3x+2y+z= 7
----------------------
2z=-6----->z=-3
since we already chose equation 1 and 3 we must involve equation 2
substituting the value if z in the second equation
5x+5y-12=3
5x+5y=15
choosing either the first or the third
3x+2y-3=7
3x+2y=10
solving the system
5x+5y=15
3x+2y=10
multipling the first equation by 2 and the second equation by 5
10x+10y=30
15x+10y=50
subtracting the two equations
-5x=-20--->x=4 substituting for the value of x
40+10y=30--->10y=-10y--->y=-1
so our soultion is x=4,y=-1,z==-3 or (4,-1,-3)








3 0
3 years ago
Can someone please help me
Zepler [3.9K]

Answer:

y=89

Step-by-step explanation:

because if you take y-47 okay then you do 47+42 giving you what that ange is because they are the same angle they should both be 42

7 0
3 years ago
The mean amount purchased by a typical customer at Churchill's Grocery Store is $27.50 with a standard deviation of $7.00. Assum
Schach [20]

Answer:

a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.

b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00

c) 90% of sample means will occur between $26.1 and $28.9.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 27.50, \sigma = 7, n = 68, s = \frac{7}{\sqrt{68}} = 0.85

a. What is the likelihood the sample mean is at least $30.00?

This is 1 subtracted by the pvalue of Z when X = 30. So

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

By the Central Limit Theorem, we have that:

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

Z = \frac{30 - 27.5}{0.85}

Z = 2.94

Z = 2.94 has a pvalue of 0.9984

1 - 0.9984 = 0.0016

0.0016 = 0.16% probability that the sample mean is at least $30.00.

b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?

This is the pvalue of Z when X = 30 subtracted by the pvalue of Z when X = 26.50. So

From a, when X = 30, Z has a pvalue of 0.9984

When X = 26.5

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

Z = \frac{26.5 - 27.5}{0.85}

Z = -1.18

Z = -1.18 has a pvalue of 0.1190

0.9984 - 0.1190 = 0.8794

0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00.

c. Within what limits will 90 percent of the sample means occur?

Between the 50 - (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile, that is, Z between -1.645 and Z = 1.645

Lower bound:

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

-1.645 = \frac{X - 27.5}{0.85}

X - 27.5 = -1.645*0.85

X = 26.1

Upper Bound:

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

1.645 = \frac{X - 27.5}{0.85}

X - 27.5 = 1.645*0.85

X = 28.9

90% of sample means will occur between $26.1 and $28.9.

4 0
3 years ago
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