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Aleonysh [2.5K]
3 years ago
9

A sum of money amounting to $3.70 consists of dimes and quarters. If there are 19 coins in all, how many are quarters?

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
4 0
Maybe 4 are quarters, im not actually sure
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Solve u^2 = -4, where u is a real number. Simplify your answer as much as possible.
alexdok [17]

Answer:

2i and -2i

Step-by-step explanation:

I'll change u to x to be easier to understand.

x^2 = -4

sqrt(x^2) = sqrt(-4)

sqrt(-4) = 2i, or -2i

8 0
2 years ago
A new bank customer with $3,500 wants to open a money market account. The bank is offering a simple interest of 1.6%.
Bond [772]

Answer:

A is $490 and B is 3990

Step-by-step explanation:

The way I solved this is with my last brain cells, you welcome.

4 0
3 years ago
What is x equal to ? Link down below.
scZoUnD [109]

Answer:

50

Step-by-step explanation:

Because of supplementary angles that line has to equal to 180

180-75 = 105

so 105 = (2x + 5)

you can subtract 5 from each side and get

100 = (2x)

then divide each side by 2 and you get

x = 50

you can check it by doing (2 x 50 + 5)

2 x 50 = 100 + 5 = 105 + 75 = 180

5 0
3 years ago
Read 2 more answers
Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg
Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let <em>X</em> = number of college women on a diet.

The probability of a woman being on diet is, P (X) = <em>p</em> = 0.70.

The sample of women selected is, <em>n</em> = 267.

The random variable thus follows a Binomial distribution with parameters <em>n</em> = 267 and <em>p</em> = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions (\hat p) follows a Normal distribution.  

The mean of this distribution is:

\mu_{\hat p} = p = 0.70

The standard deviation of this distribution is: \sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z

**Use the <em>z</em>-table for the probability.

P(\hat p \geq 0.75)=1-P(Z

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

7 0
3 years ago
Using radicals, what is an equivalent expression for the expression 2 1/3
faltersainse [42]

Answer:

7

Step-by-step explanation:

21/3

3 will council it self one and 21 7times

4 0
3 years ago
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