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Lesechka [4]
3 years ago
6

You are going to mix a gallon bucket of window cleaner. The instructions direct you to mix 1 part cleaner to 3 parts water. How

much cleaning solution will you need to use?
A) 1 ½ quarts B)1 pint C) 2 quarts D) 1 quart
Mathematics
2 answers:
Pie3 years ago
4 0
One gallon equals to 4 quarts.

The instructions tell you to add 1 part of cleaner and 3 parts of water, in total 4 parts. It makes it easy as you can take that one part = one quart.

You will use 3 quarts of water and 1 quart of cleaner. The answer is D.


Natalka [10]3 years ago
3 0
Let
x--------> gallons of cleaner
y-------> gallons of water

we know that
x+y=1 equation 1
and
\frac{x}{y} = \frac{1}{3}  \\ y=3x equation 2
substitute equation 2 in equation 1
x+3x=1 \\ 4x=1 \\ x=1/4 gal

find the value of y
y=3x \\ y=3* \frac{1}{4}  \\ y= \frac{3}{4} gal

we know that
\frac{1}{4} gal is equal to 1 quart

therefore
the answer is the option
<span>D) 1 quart</span>
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Suppose we take a random sample of 41 state college students. Then we measure the length of their right foot in centimeters. We
Debora [2.8K]

Answer:

ME = \frac{25.09-21.01}{2}= 1.69

The general formula for the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

And for this case the width is:

Width= 2*t_{\alpha/2} \frac{s}{\sqrt{n}}

And if we decrease the confidence level from 95% to 90% then the critical value t_{\alpha/2} would decrease and in effect the width for this new confidence interval decreases.

As confidence level decreases, the interval width decreases

Step-by-step explanation:

For this cae we know that the sample size selected is n =41

And we have a confidence interva for the true mean of foot length for students at a college selected.

The confidence interval is given by this formula:

\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}

And for this case the 95% confidence interval is given by: (21.71,25.09)

A point of etimate for the true mean is given by:

\bar X = \frac{21.71+25.09}{2}= 23.4

And the margin of error would be:

ME = \frac{25.09-21.01}{2}= 1.69

The general formula for the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

And for this case the width is:

Width= 2*t_{\alpha/2} \frac{s}{\sqrt{n}}

And if we decrease the confidence level from 95% to 90% then the critical value t_{\alpha/2} would decrease and in effect the width for this new confidence interval decreases.

As confidence level decreases, the interval width decreases

3 0
3 years ago
Can anyone do these problems it’s for my daughter in law
Naya [18.7K]

Classwork:

Given f(x) = x^2 - 1 and g(x)=2x+5, we have

(1) (f\circ g)(x) = f(g(x)) = f(2x+5) = (2x+5)^2 - 1 = \boxed{4x^2+20x+24}

Using the composition found in (1), we have

(2) (f\circ g)(-2) = 4\cdot(-2)^2+20\cdot(-2)+24 = \boxed{0}

(3) (g\circ f)(x) = g(f(x)) = g(x^2-1) = 2(x^2-1) + 5 = \boxed{2x^2 + 3}

Using the composition found in (3),

(4) (g\circ f)(1) = 2\cdot1^2+3 = \boxed{5}

Homework:

Now if f(x)=x^2-3x+2, we would have

(1) (f\circ g)(x) = f(2x+5) = (2x+5)^2-3(2x+5)+2 = \boxed{4x^2+14x+12}

For (2), we could explicitly find (g\circ f)(x) then evaluate it at <em>x</em> = -1 like we did in the classwork section, but we don't need to.

(2) (g\circ f)(-1) = g(f(-1)) = g((-1)^2-3\cdot(-1)+2) = g(6) = 2\cdot6+5 = \boxed{17}

(3) We can demonstrate that both methods work here:

• by using the result from (1),

(f\circ g)(2) = 4\cdot2^2+14\cdot2+12 = \boxed{56}

• by evaluating the inner function at <em>x</em> = 2 first,

(f\circ g)(2) = f(g(2)) = f(2\cdot2+5) = f(9) = 9^2-3\cdot9+2 = \boxed{56}

4 0
2 years ago
Choose the required figure.
zloy xaker [14]

not sure how the first two pictures apply here, since they're just a single circle.


check the picture below.


the one on the left can have 3 common tangent lines, the one on the right, only two common ones.

5 0
3 years ago
Plzzz help me with this I’ll give brainliest
scoray [572]

Answer18:

The quadrilateral ABCD is not a parallelogram

Answer19:

The quadrilateral ABCD is a parallelogram

Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

The slope of a line is given m=\frac{Y2-Y1}{X2-X1}

Now,

The slope of a line AB:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{6-(-1)}{(-4)-(-4)}

m=\frac{7}{0}

The slope is 90 degree

The slope of a line BC:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{6-6}{(-4)-(-1)}

m=\frac{0}{(-3)}

The slope is zero degree

The slope of a line CD:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-4)-6}{2-2}

m=\frac{-10}{0}

The slope is 90 degree

The slope of a line DA:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-1)-(-4)}{(-4)-(2)}

m=\frac{3}{-6}

m=\frac{-1}{2}

The slope of the only line AB and CD are the same.

Thus, The quadrilateral ABCD is not a parallelogram

For question 19:

Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

The slope of a line is given m=\frac{Y2-Y1}{X2-X1}

Now,

The slope of a line AB:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{2-3}{3-(-2)}

m=\frac{-1}{5}

The slope of a line BC:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-1)-2}{2-3}

m=\frac{-3}{-1}

m=3

The slope of a line CD:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{0-(-1)}{(-3)-2}

m=\frac{-1}{5}

The slope of a line DA:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{3-0}{(-2)-(-3)}

m=3

The slope of the line AB and CD are the same

The slope of the line BC and DA are the same

Thus, The quadrilateral ABCD is a parallelogram

3 0
3 years ago
The current value of an investment is 25% more than its initial value. The increase in value is $10 million. What was the initia
katrin [286]
B is the answer ////////
7 0
3 years ago
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