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laila [671]
3 years ago
6

The math club is having a car wash to raise funds to attend a state competition. They need to raise a minimum of $325.00. The co

st of supplies for the car wash is $25.00. If they expect 40 cars on the day of the car wash, what is the amount they will have to charge each car to meet the minimum goal?
Mathematics
2 answers:
Svetlanka [38]3 years ago
7 0

Answer:

8.75

Step-by-step explanation:

the formula to use would be:

$325 = 40x - 25

add 25 to both sides of the equation to get:

$350 = 40x

divide both sides by 40 to get:

x = 350/40 = $8.75

if they charge $8.75 per car, they will raise 40 * that for a total of $350.

they would then subtract the $25 for the cost of the supplies and they would be left with $325.

that's their minimum goal.

777dan777 [17]3 years ago
6 0

Answer:

8.75

Step-by-step explanation:

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Suppose a chemist combines a 25% acid solution and a 50% acid solution to make 40 L of 45% acid solution. How many liters of eac
lesantik [10]

Answer:

The number of liters for :

Acid solution a = x = 8 liters

Acid solution b = y = 32 liters

Step-by-step explanation:

Let us represent:

The number of liters for :

Acid solution a = x

Acid solution b = y

Suppose a chemist combines a 25% acid solution and a 50% acid solution to make 40 L of 45% acid solution.

x + y = 40 ...... Equation 1

x = 40 - y

25% × x + 50% × y = 45% × 40

0.25x + 0.5y = 18...... Equation 2

We substitute, 40 - y for x in Equation 2

0.25(40 - y)+ 0.5y = 18

10 - 0.25y + 0.5y = 18

- 0.25y + 0.5y = 18 - 10

0.25y = 8

y = 8/0.25

y = 32 Liters

Solving for x

x = 40 - y

x = 40 - 32

x = 8 Liters.

Hence:

The number of liters for :

Acid solution a = x = 8 liters

Acid solution b = y = 32 liters

5 0
3 years ago
We wish to give a 90% confidence interval for the mean value of a normally distributed random variable. We obtain a simple rando
coldgirl [10]

Answer: (9.27025,\ 11.12975)

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean : \overline{x}=10.2

sample standard deviation : s= 1.5

Significance level ; \alpha= 1-0.90=0.10

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value = t_{\alpha/2, df}=t_{0.05 , 8}=1.8595

Confidence interval for the population mean :

\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}

90% confidence interval for the mean value will be :

10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}

10.2\pm (1.8595)\dfrac{1.5}{3}

10.2\pm (1.8595)(0.5)

10.2\pm (0.92975)

(10.2-0.92975,\ 10.2+0.92975)

(9.27025,\ 11.12975)

Hence, the 90% confidence interval for the mean value= (9.27025,\ 11.12975)

6 0
3 years ago
The mass of an ant is 1/10 of the load, which it can carry in one try. What is the mass of the ant, if in one time it can carry
Viefleur [7K]

If you call m the mass of the ant and l the load, we have the equation

m = \cfrac{l}{10}

In fact, the mass of the ant is one tenth of the load, which is exactly what this equation states.

Since we are given the load, we simply need to plug its value in the equation to deduce the mass of the ant:

m = \cfrac{\frac{7}{250}}{10} = \cfrac{7}{250}\cdot\cfrac{1}{10} = \cfrac{7}{2500}

8 0
4 years ago
I hate this pain i got through i just cant anymore...<br><br> y+50=700
DiKsa [7]

Answer:

y=650

Step-by-step explanation:

Just subtract 50 from each side

And Yes, please

5 0
3 years ago
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A sandbox has a perimeter of 21 1/2 feet. If the length is 5 1/4 feet, what is the area of the sand box?
LiRa [457]
The perimeter of a rectangle is twice the sum of length and width, so the sum of length and width is half the perimeter, 10 3/4 ft.

The width will be the difference between this and the length, hence 5 1/2 ft.

The area is the product of length and width.
  A = length×width
  A = (5.25 ft)×(5.5 ft) = 28.875 ft²

The area of the sandbox is 28 7/8 ft².
4 0
3 years ago
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