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11Alexandr11 [23.1K]
3 years ago
6

I need help plssssss plsss here seggy picture too

Mathematics
2 answers:
Mrac [35]3 years ago
6 0

is it froom fairy tale

Step-by-step explanation:

k your welcome

Novay_Z [31]3 years ago
3 0

Answer:

So seggzsey

Step-by-step explanation:

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Can someone help me with this?
Wewaii [24]
<span>3x+4=y+5, then y+5=3x+4
It is called: reflexive property

-----------------------------
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<span>x = y and y = 10, then x = 10
It is called: equality substitution property  
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7 0
3 years ago
The hiski s family spends $600 per month on groceries for the family. If this is 15% of their monthly budget , then what is thei
ad-work [718]

I think that it is $4,000

8 0
4 years ago
REAL ANSWERS PLEASE AND NOT JUST FOR THE POINTS 20 EACH
Fed [463]
A+b=(-3n+2)+(5n-7)

=(5-3n)+(2-7)

=2n-5


Came up with the same answer as the first guy
4 0
3 years ago
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
Solve the following equations.<br> 43xx−1 = 32
kotykmax [81]
This is the answer.

3 0
3 years ago
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