Ask question

Ask question

All categories

3 years ago

13

Bill cut out 6 squares and 18 circles . He divided the cutouts into groups so that the same number of squares and circles were i

n each groups . What is the greatest number of groups he could have made ?

2 answers:

Tcecarenko [31]3 years ago

6 0

The greatest number of groups he could have made are 3

sertanlavr [38]3 years ago

5 0

The greatest number of groups he could have made is three. 2 squares and 6 circles in each group.

You might be interested in

Judy is a single woman, 25 years old. Her take home pay is $2314.92/month. She also earns $48 interest per month on her savings

Leni [432]

The monthly budget plan for Judy includes all of her income and expenses, the total balance is $ 2362.92 and she spends $ 1083.

<h3>What is the monthly budget?</h3>

A monthly budget is a personal spending plan that you use to track your monthly income and costs.

The given data in the problem is;

Home pay = $2314.92/month.

Interest earned = $48 / month

House rent = $825/month

EMI of student loan = $258/month

Total earning per month = $2314.92 + $48

Total earning per month = $ 2362.92

Amount she spends per month = $825/month + $258

Amount she spends per month = $ 1083

Hence, the total balance will be $ 2362.92 and she spend $ 1083.

To learn more about the monthly budget refer to:

brainly.com/question/570904

#SPJ1

3 0

1 year ago

The set of ordered pairs represented by the graph below can be described as which of the following

Aneli [31]

ANSWER

a relation only

EXPLANATION

The given graph shown in the attachment represents only a relation and not a function.

The reason is that, a vertical line drawn across this graph will intersect this graph at more than one point.

Since the graph fails to pass the vertical line test, the ordered pair represented by this graph represents a relation only.

6 0

3 years ago

Read 2 more answers

36√(3)-(432/7+4√(3))=?

Elden [556K]

40√3+432/7 is the answer

Hope this helps :D

5 0

3 years ago

The solution to the system equations

katovenus [111]

Let's solve this system of equations through substitution.

We have these two equations.

-7x-2y=14

6x+6y=18

Now let divide the second equation by 6.

6x+6y=18 ----> x+y=3

Next, let us move y to the right side of the equation.

x+y=3 -------> x=3-y (x equals 3-y)

Because we found out that what x is in terms of y, we can input that in for every instance of x in this equation below.

-7x-2y=14 becomes -7(3-y)-2y=14 (Why? Because x equals 3-y!)

We have a one variable equation now and can solve for y.

-7(3-y)-2y=14

-21+7y-2y=14

5y=35

y=7

Plug in 7 for y in any equation to find x.

x+y=3

x+7=3

x=-4

answer: x=-4, y=7

3 0

2 years ago

The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are

almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = <u><em>the number that are not correctly calibrated.</em></u>

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_3$

So, the required probability = $\frac{^{6}C_3}{^{8}C_3}$

= $\frac{20}{56}$ = <u>0.35</u>

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_2$

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_1$

So, the required probability = $\frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}$

= $\frac{30}{56}$ = <u>0.54</u>

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_1$

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_2$

So, the required probability = $\frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}$

= $\frac{6}{56}$ = <u>0.11</u>

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0

3 years ago