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saw5 [17]
3 years ago
10

Richard has four times as many marbles as john. if Richard have 18 to John they would have the same number. how many marbles has

each?​
Mathematics
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

The number of marbles Richard has 24 and John has 6.

Step-by-step explanation:

Richard has four times as many marbles as john.

If Richard have 18 to John they would have the same number.

Now, to find number of marbles each has.

Let the marbles of John be x.

And the marbles of Richard be 4x.

According to question:

4x=x+18.

<em>Subtracting both sides by x we get:</em>

3x=18.

<em>Dividing both sides by 3 we get:</em>

x=6.

<u><em>Marbles of John</em></u><em> = 6.</em>

<u><em>Marbles of Richard</em></u><em> = 4x=4\times 6=24</em>

Therefore, the number of marbles Richard has 24 and John has 6.

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4. East High School has 540 students. There are 220 girls in the school.
Ierofanga [76]

Answer:

11:16

Step-by-step explanation:

The ratio is 220:320.

Cancel the 0s then divide by 2.

22/2=11

32/2=16

3 0
3 years ago
2+2×4 that is my question i need help​
devlian [24]

Answer:

10

Step-by-step explanation:

2+(2x4)

2x4=8

2+8=10

6 0
3 years ago
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What is the slope of y=2x-6
Korvikt [17]
The slope is 2x. In other words 2/1
5 0
3 years ago
Whats 1500x1500+(1/5- 2/6) divided by 28=?
Ilya [14]

Answer:

2249999.99524

Step-by-step explanation:

The answer would be 2249999.99524

Or just 224999

Sorry for the last answer!

5 0
2 years ago
An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results an
igomit [66]

Answer:

Mean = 1.57

Variance=0.31

Step-by-step explanation:

To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:

1) Both surgeries are successful (X=2).

P(e_1)=0.90*0.67=0.603

2) Left knee unsuccessful and right knee successful (X=1).

P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067

3) Right knee unsuccessful and left knee successful (X=1).

P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297

4) Both surgeries are unsuccessful (X=0).

P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033

Then, the mean can be calculated as the expected value:

M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57

The variance can be calculated as:

V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111

3 0
3 years ago
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