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Dovator [93]
2 years ago
10

A lotion is made from an oil blend costing $1.50 per ounce and glycerin costing $1.00 per ounce. Four ounces of lotion costs $5.

50.
A table titled Lotion, showing Ounces, Cost per Ounce, and Total. The first row shows, Oil Blend, and has 4 minus g, 1.50, and x. The second row shows Glycerin, and has g, 1.00, and blank. The third row shows, Mixture, and has 4, blank, and 5.50.

Which value could replace x on the table?

(4 – g) 1.5
(4 – g) + 1.5
4g – g(1.5)
(4 – g)g
Mathematics
1 answer:
Svet_ta [14]2 years ago
8 0

Answer:

A (4-g)1.5

Step-by-step explanation:

hope this helps if you need a explanation just ask in the comments and I will.

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Answer:

1) -32+12x 2) 2(3x-4)

Step-by-step explanation:

I hope this is right!

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7th GRADE MATH WILL GIVE BRAINLIEST 20 POINTS ​
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Answer:

I will say C, because tossing a coin twice has two outcomes. H is for heads while T is for heads. Tell me if I'm right.

Step-by-step explanation:

4 0
3 years ago
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amm1812
Where is the question and what is it?
4 0
2 years ago
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Find the slope of the line that passes through (9, 6) and (4, 5).
boyakko [2]

Answer:

slope = \frac{1}{5}

Step-by-step explanation:

calculate the slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (9, 6 ) and (x₂, y₂ ) = (4, 5 )

m = \frac{5-6}{4-9} = \frac{-1}{-5} = \frac{1}{5}

8 0
2 years ago
A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
2 years ago
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