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nydimaria [60]
3 years ago
10

This is simple, but I don't have time to do this now, list the composite numbers from 2-50 please!

Mathematics
1 answer:
Kamila [148]3 years ago
6 0
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50
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A school sold tickets to a musical. The school received $6.50 per ticket sold.
Kamila [148]

Answer: m=$6.50t AND t$6.50=m

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!!
Dafna1 [17]

Answer:

Step-by-step explanation:

Any time you have compounding more than once a year (which is annually), unless we are talking about compounding continuously, you will use the formula

A(t)=P(1+\frac{r}{n})^{(n)(t)}

Here's what we have:

The amount after a certain time that she has in the bank is 4672.12; that's A(t).

The interest rate in decimal form is .18; that's r.

The number of times the interest compounds is 12; that's n

and the time that the money is invested is 3.5 years; that's t.

Filling all that into the formula:

4672.12=P(1+\frac{.18}{12})^{(12)(3.5)} Simplifying it down a bit:

4672.12=P(1.015)^{42} Raise 1.015 to the 42nd power to get

4672.12 = P(1.868847115) and divide to get P alone:

P = 2500.00

She invested $2500.00 initially.

6 0
3 years ago
The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to d
NikAS [45]

Answer:

Option b. should not be rejected

Step-by-step explanation:

We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.

We have to test whether the variance of the population is significantly more than 0.003, i.e.;

  Null Hypothesis, H_0 : \sigma = \sqrt{0.003}

Alternate Hypothesis, H_1 : \sigma > \sqrt{0.003}

The test statistics used here for testing variance is;

          T.S. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2}__n_-_1

where, s = sample standard deviation = 0.06

           n = sample size = 26 cans

So, Test statistics = \frac{(26-1)0.06^{2} }{0.003 } ~ \chi^{2}__2_5

                            = 30

So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.

5 0
3 years ago
I need help on this, please hurry and thank you!
inna [77]

Answer:

A.\ a_n=n^2+1

Step-by-step explanation:

Check:

a_n=n^2+1\\\\a_1=1^2+1=1+1=2\qquad CORRECT\\\\a_2=2^2+1=4+1=5\qquad CORRECT\\\\a_3=3^2+1=9+1=10\qquad CORRECT\\\\a_4=4^2+1=16+1=17\qquad CORRECT\\\\a_5=5^2+1=25+1=26\qquad CORRECT

Used PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

First Power, next Addition

4 0
3 years ago
Read 2 more answers
Line A is perpendicular to Line B.
Darya [45]

Answer:

-7/3

Step-by-step explanation:

the answer is the negative reciprocal.

6 0
3 years ago
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