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Step2247 [10]
3 years ago
8

Ming has $7:35 in quarters and nickels. She had a total of 79 coins. How many quarters and how many nickels does she have

Mathematics
2 answers:
jasenka [17]3 years ago
7 0
7×4=28
35×1=35
I think Ming has 28 quarters and 35 nickels.
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

62 nickels and 17 quarters

Step-by-step explanation:

62x0.5=3.10

17x0.25=4.25

4.25+3.10=7.35

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The diameter of a semicircle is 6 miles. What is the semicircle's perimeter?
koban [17]

Answer:

15.42mi

Step-by-step explanation:

Use the formula for a semicircle's perimeter.

P=\frac{1}{2}\pi *d+d

Plug in 6 for d.

P=\frac{1}{2}\pi*6+6

Let's use 3.14 for \pi, just to make it easier, but of course, if it states to round it to something else, just plug in that many values for \pi.

P=\frac{1}{2}(3.14)*6+6 \\ \\ P=1.57*6+6 \\ \\ P=9.42+6 \\ \\ P=15.42

5 0
3 years ago
Read 2 more answers
allen has a debt of 500$.He owes equal amounts to 10 different people. Express his debt to one person as a negative number.
SOVA2 [1]
-50 because if he owes 10 different people the same amount it would be -50 because he owes them money
4 0
2 years ago
Graph 3(8-4x)<6(x-5)
Zinaida [17]

First simplify this inequality:

3(8-4x)

Now you can graph this inequality. Draw the vertical dashed line x=3 and shade the region to the right from this line. This is exactly the region that represents the solution of inequality (see attached diagram for details).

4 0
3 years ago
A Martian couple has children until they have 2 males (sexes of children are independent). Compute the expected number of childr
Ne4ueva [31]

Answer:

a) 6

b) 4

c) 3

Step-by-step explanation:

Let p be the probability of having a female Martian, and of course, 1-p the probability of having a male Martian.

To compute the expected total number of trials before 2 males are born, imagine an experiment simulating the fact that 2 males are born is performed n times.

Let ak be the number of trials performed until 2 males are born in experiment k. That is,

a1= number of trials performed until 2 males are born in experiment 1

a2= number of trials performed until 2 males are born in experiment 2

and so on.

If a1 + a2 + … + an = N

we would expect Np females.  

Since the experiment was performed n times, there 2n males (recall that the experiment stops when 2 males are born).

So we would expect 2n = N(1-p), or

N/n = 2/(1-p)

But N/n is the average number of trials per experiment, that is, the expectation.

<em>We have then that the expected number of trials before 2 males are born is 2/(1-p) where p is the probability of having a female. </em>

a)

Here we have the probability of having a male is half as likely as females. So

1-p = p/2 hence p=2/3

The expected number of trials would be

2/(1-2/3) = 2/(1/3) =6

This means <em>the couple would have 6 children</em>: 4 females (the first 4 trials) and 2 males (the last 2 trials).

b)

Here the probability of having a female = probability of having a male = 1/2

The expected number of trials would be

2/(1/2) = 4

This means<em> the couple would have 4 children</em>: 2 females (the first 2 trials) and 2 males (the last 2 trials).

c)

Here, 1-p = 2p so p=1/3

The expected number of trials would be

2/(1-1/3) = 2/(2/3) = 6/2 =3

This means<em> the couple would have 3 children</em>: 1 female (the first trial) and 2 males (the last 2 trials).

5 0
3 years ago
How would you write "there were 3 times as many baby dinosaurs as adult dinosaurs" as an algebraic equation where x = baby dinos
Mrrafil [7]
X = 3y

hope it helped!
5 0
2 years ago
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