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3 years ago

What is 2+2 will mark brainliest

Mathematics

1 answer:

sasho [114]3 years ago

4 0

Answer:

It 4 and thank you

Step-by-step explanation:

You might be interested in

How do you turn 584.56714755 into a radical answer and what would it be

Llana [10]

Answer

584.56714755 = 146

Step-by-step explanation:

584.567 is 4 times 146 ...

...and 4 can be written as 2 to the power 2.

So, the 4 can just be pulled outside of the radical sign, as a 2 ,

and the other factor, the 146 , just stays inside of the radical sign.

8 0

3 years ago

there are about 25000000000 red boold cells in the average adult. About how many adult would it take to have a total googol of r

IRISSAK [1]

A googol is 1.0*10^100. there are 25000000000 RBC in the average adult.
The number of adults equals
1.0*10^100 / 25000000000
=400000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
(=10^100/(25*10^9)=4*10^89)

6 0

3 years ago

A ride at a theme park lasts on and one half minutes.it takes 2 minutes to prepare the ride for each trip how many times can the

nlexa [21]

1 time=2mins
xtimes=30mins

30/2 ×1= 15.
Therefore it can be completed 15 times in 30mins

6 0

3 years ago

Read 2 more answers

20 POINTS IF YOU SOLVE!

Orlov [11]

Question 1: D

Question 2:D

Question 3: C

Question 4: A

Question 5: D

5 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