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k0ka [10]
3 years ago
14

13.25 x 64 please awnser it's for a test and I have no idea what it is. ​

Mathematics
2 answers:
balu736 [363]3 years ago
5 0

Answer:

453x  ^64 /4

​

Step-by-step explanation:

stepladder [879]3 years ago
3 0

Answer :

13.25 x 64

=848

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the 2nd term in a geometric sequence is 20 the 4th term in the same sequence is 45/4 or 11.25 what is the common ratio?
lbvjy [14]
Let x be the common ratio and y be the 3rd term

11.25/x=y     This means the fourth term divided by the common ratio will get the third term
20x=y           This means the second term multiplied by the common ratio will get the third term

11.25/x=20x        Set the two y's equal to each other, and solve for x!
20x^2=11.25
x^2=.5625
x=+-√.5625
x=.75 or -.75
5 0
3 years ago
Read 2 more answers
Find the balance of $7,000 deposited at 4% compounded semi-annually for 2 years
Rufina [12.5K]

Answer:

The balance will be $7,577.03.

Step-by-step explanation:

The compound interest formula is given by:

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

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this problem, we have that:

P = 7000, r = 0.04

Semianually means twice a year, so n = 2

We want to find A when t = 2.

So

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

A = 7000(1 + \frac{0.04}{2})^{2*2}

A = 7577.03

The balance will be $7,577.03.

5 0
3 years ago
Write the number 4<br> 4 as a product of prime factors.
Paladinen [302]
The answer would be
 2 x 2 = 4

7 0
3 years ago
In a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50. Determ
shepuryov [24]

Answer:

t-distribution should be used to construct a confidence interval.

Step-by-step explanation:

We are given that a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50.

We have to determine whether a normal distribution (Z values) or a t- distribution should be used or whether neither of these can be used to construct a confidence interval.

<em><u>Since in this question we are provided with;</u></em>

Sample average weekly food expense, \bar X = $95.60

Sample standard deviation, s = $22.50

Sample of families, n = 18

The distribution that we will use here to construct a confidence interval will be <u>t-distribution</u> because in the question we don't know anything about population standard deviation (\sigma) .

Normal distribution is used when we know population standard deviation (\sigma).

So, the pivotal quantity for confidence interval that will be used is One-sample t-test statistics;

                P.Q. = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

Therefore, t-distribution should be used to construct a confidence interval.

3 0
3 years ago
Convert to a mixed number, reduce when possible (-3.4 x -6.6)
julia-pushkina [17]

The answer would be 22,44

3 0
3 years ago
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