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ohaa [14]
1 year ago
14

Solve the system of equations

Mathematics
2 answers:
Illusion [34]1 year ago
5 0

Answer:

x=-8, y=2

Step-by-step explanation:

if 1/2x+6y=12 then 1/2x+y=2 and x+y=1 and y=1-x..if y=1-x then 1-x=x=15 and 2x=16 so x=8 if x=8 then 1/2(8) +6y=12 or 4=6y=12.. then 2y=4 and y=2

kaheart [24]1 year ago
3 0

Answer:

x=-12,y=3

Step-by-step explanation:

1/2x+6y=12

y=x+15

since y=x+15

substitute y in first equation, we have

1/2x+6(x+15)=12

open the bracket

1/2x+6x+90=12

1/2x+6x=12-90

x/2+6x= -78

13x/2=-78

13x=2×-78

x=-12

Now x=-12, put x=-12 in the second equation, we have

y=x+15

y=-12+15

y=3

hope it helps

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Blank numbers can be written as a product of blank factors.This is called the prime factorization of a number.
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Given g(x) = 2x + 3, find g(3).
Kaylis [27]

Answer:

g(3) = 9

Step-by-step explanation:

replace 3 in x

g(3) = 2(3) +3 = 6 + 3 = 9

Hope this helps

7 0
10 months ago
Add a kennel there are five cats and six dogs what are the correct ratios
hodyreva [135]

Answer:

6:5

Step-by-step explanation:

I am thinking it would be 6:5 because you have 6 dogs, and 5 cats. So, you could think of it as for every 6 dogs you have 5 cats at the kennel.

6 0
2 years ago
A university financial aid office polled a random sample of 670 male undergraduate students and 617 female undergraduate student
ahrayia [7]

Answer:

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Male undergraduates:

670, 388 were employed. So

p_M = \frac{388}{670} = 0.5791

s_M = \sqrt{\frac{0.5791*0.4209}{670}} = 0.0191

Female undergraduates:

Of 617, 323 were employed. So

p_F = \frac{323}{617} = 0.5235

s_F = \sqrt{\frac{0.5235*0.4765}{617}} = 0.0201

Distribution of the difference:

p = p_M - p_F = 0.5791 - 0.5235 = 0.0556

s = sqrt{s_M^2+s_F^2} = \sqrt{0.0201^2 + 0.0191^2} = 0.0277

Confidence interval:

The confidence interval is given by:

p \pm zs

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The lower bound of the interval is:

p - zs = 0.0556 - 1.645*0.0277 = 0.01

The upper bound of the interval is:

p + zs = 0.0556 + 1.645*0.0277 = 0.1012

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

5 0
3 years ago
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