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Misha Larkins [42]
2 years ago
11

By selling 80 oranges a shopkeeper loses the cost price of 20 oranges.Find his loss percent.​

Mathematics
1 answer:
tatuchka [14]2 years ago
6 0

Answer:25%

Step-by-step explanation:

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PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA 40 POINTS* DONT SKIP :(( .!
Juli2301 [7.4K]

Answer:

QPN and TSU

Step-by-step explanation:

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3 years ago
Factor please <br> x^2+8x+15
Anastasy [175]

Answer:

(x + 3)(x + 5)

Step-by-step explanation:

The given expression is x^2 +8x + 15

Now find the factor of 15 such that when you add the factors, we have to get 8

The right factors of 15.

5 * 3 = 15

5 + 3 = 8

The right factors are 3 and 5.

x^2 + 8x + 15 = (x + 3)(x + 5)

Therefore, x^2 + 8x + 15 = (x + 3)(x + 5)

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3 years ago
One of the problems with our current election system is??
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Answer:

Assuming you mean the US election a problem with our electron is the person who wins the most popular vote doesn't always win.

Why is this a problem?

We vote for who we want as a president as it should work, but then the most elected doesn't always become president. Almost like our votes do not matter and is very discouraging.

Hope this helps!

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3 years ago
PLEASE PLEASE HELP ME!
Paladinen [302]
The answer is D
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4 years ago
Read 2 more answers
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
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