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Black_prince [1.1K]
3 years ago
11

I know how to do this, but I have so much homework soo please help

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
4 0
What’s ur question tho?
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Who can help me with this for algebra 2
Anuta_ua [19.1K]
The answer to this equation is

x= 1, -3
7 0
3 years ago
Round 37.8 to the nearest whole number.
STALIN [3.7K]
Any decimal lower that .5 is rounded down, and anything .5 or higher is rounded up, so the answer would be 38.


3 0
4 years ago
Read 2 more answers
What is a co-prime number? Please answer as soon as possible Thank you
ipn [44]

Answer:

\boxed{\mathrm{view \: explanation}}

Step-by-step explanation:

Two numbers that only have 1 as a common factor are called co-prime numbers.

<u>Example:</u>

Factors of 3 ⇒ 1, 3

Factors of 4 ⇒ 1, 2, 4

These numbers only have 1 as a common factor. So 3 and 4 are co-prime numbers.

6 0
3 years ago
A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of
GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.989.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

8 0
3 years ago
NEED HELP ASAP PLEASE
Tju [1.3M]

Answer:

9/35

Step-by-step explanation:

no of even number =3(2,4,6)

probablity of getting even in both dice =(3/6)×(3/6)

=9/36

4 0
3 years ago
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