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galben [10]
3 years ago
6

one tamk of gasoline has an octane rating of 140 and another tank of gasoline has an octane rating of 80. To obtain a mixture of

60 gallons with an octane rating of 120, how many gallons should be used from the tank with the octane rating of 80?
Mathematics
1 answer:
Zarrin [17]3 years ago
7 0

Answer:

20 gallons were used from the tank with the octane rating of 80.

Step-by-step explanation:

Given : One tank of gasoline has an octane rating of 140 and another tank of gasoline has an octane rating of 80. To obtain a mixture of 60 gallons with an octane rating of 120.

To find : How many gallons should be used from the tank with the octane rating of 80?

Solution : Let x be the volume of gas whose octane rating is 140 and

y be the volume of gas whose octane rating is 80.

Then, we form the equation as:

(1) x\times140 + y\times80 = 60\times120

140x+80y = 7200

or  7x+4y = 360 .........[3]

(2) x + y = 60

Now multiply equation (2) by 4  

4x + 4y = 240 ...........[4]

Subtract (4) from (3)  

7x+4y-4x-4y=360-240

3x=120

x=\frac{120}{3}

x=40

Put x in equation [4] we get,

4(40) + 4y = 240

160 + 4y = 240

4y = 80

y=\frac{80}{4}

y=20

Therefore, x=40 and y=20

So, 20 gallons were used from the tank with the octane rating of 80.






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posledela

Answer:

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4 0
3 years ago
A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th
ivolga24 [154]

Answer:

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

H_0: \mu=500\\\\H_a:\mu< 500

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1

The degrees of freedom for this sample size are:

df=n-1=5-1=4

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

4 0
3 years ago
Jaylynn was buying new mascara. She bought it on sale for $5.56. If the price represents a 20% discount from the original price,
Len [333]
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4 0
3 years ago
PLEASE HELP!
Kitty [74]
You can reword the two equations as:

-5x-y=15 (Divide original value by 3)
-2x+6y=6

Then use elimination to find x:

-30x-6y=90 (Multiply by 6 to get y values to be same to cancel out)
-2x+6y=6

You're left with:

-32x=96. Which can then be solved to find x which is -3.

Then plug back in 

-2x+6y=6

Now to: -2(-3)+6y=6.

Which reduces to 6+6y=6. So y=0.



To graph them, just reword the equations (yes once again) so that y is in front.

y=-5x-15 and y=(1/3)x+1






8 0
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