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Andrews [41]
2 years ago
12

(PLEASE ANSWER 50 POINTS)Write an equation of a line slope-intercept form with the given slope and y-intercept. Slope:3, y-inter

cept:2
Mathematics
1 answer:
Levart [38]2 years ago
8 0
Y=3x+2

the slope (mx) is 3
the y-intercept (b) is 2.

the equation form is y=mx+b
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Justine charges $25 to wash a dog. She washed 5 dogs on Saturday and then some more on Sunday. She made $325 for the weekend. Ho
Marrrta [24]

Answer

• A. Equation: 25(5 + x) = 325

,

• B. Answer: 8 dogs

Explanation

Given

• Charge to wash a dog: $25.

• She washed 5 dogs on Saturday and then some more on Sunday.

• She made $325 for the weekend.

Procedure

She charges $25 per wash, she made $325 for the weekend, and we know that on Saturday she washed 5 dogs, but we don't know how many she washed on Sunday. Thus, we have to build an equation in which the number of dogs washed on Sunday is represented by x (as we do not know the real number).

Considering that the total money made has to be equal to the multiplication of the charge times the dogs washed, the equation is:

25(5+x)=325

Then, we have to solve for x to know how many dogs did she wash on Sunday.

0. Multiplying the parenthesis

125+25x=325

<em>2. Subtracting 125 from both sides of the equation</em>

125-125+25x=325-12525x=200

<em>3. Dividing both sides of the equation against 25</em>

\frac{25x}{25}=\frac{200}{25}x=8

8 0
1 year ago
The rule for the number of fish in a home aquarium is 1 gallon of water for each inch of fish length. Marta's aquarium holds 22
Anuta_ua [19.1K]

Answer:

Fish A is 4 inches.

Fish B is 5 inches.

Step-by-step explanation:

22=3a+2b

and

32=3a+4b

So,

a=4 and b=5

7 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

7 0
3 years ago
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