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3 years ago

7

If 5 candy bars cost $3.00, how much will 2 candy bars cost?

1 answer:

denis23 [38]3 years ago

8 0

wait nvm

one candy bar costs 0.60 cents

multiply this number by 2 and get <u><em>$1.20</em></u>

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What’s is 2 1/2 divided 1/3

Marat540 [252]

Answer:

7 $\frac{1}{2}$

Step-by-step explanation:

$2\frac{1}{2} *$ the reciprocal of $\frac{1}{3}$

To find reciprocal, flip the top and the bottom.

So $\frac{1}{3}$ would be $\frac{3}{1}$ .

Multiply like normal.

6 0

3 years ago

Rudolph can fly 5 miles in 12 seconds. At<br> that rate, how many miles can he fly per<br> minute?

AleksandrR [38]

Answer:

25 mpm

Step-by-step explanation:

60 devided by 12 = 5

5 times 5 = 25 mpm

7 0

3 years ago

Find two positive consecutive odd integers such that square of the smaller integer is 10 more than the larger integer

Otrada [13]

Let $2n+1$ be the smaller integer. The larger integer is then $2n+3$ , and we have

$(2n+1)^2=10+(2n+3)\implies4n^2+4n+1=2n+13$

$\implies4n^2+2n-12=0$

$\implies2n^2+n-6=0$

$\implies(2n-3)(n+2)=0$

$\implies 2n-3=0\text{ or }n+2=0$

$\implies n=\dfrac32\text{ or }n=-2$

We omit $n=-2$ , since $2(-2)+1=-3$ is negative.

Then for $n=\dfrac32$ we find $2\left(\dfrac32\right)+1=4$ , but this is not odd.

There are no consecutive odd integers that satisfy the given condition!

3 0

3 years ago

Find the Greatest Common Factor of 15 and 20

puteri [66]

Answer:

5

Step-by-step explanation: i created the list of numbers.

15: 1,3,<u>5</u>,15

20: 1,2,4,<u>5</u>,10,20

can i get branliest

4 0

3 years ago

Coach Evans recorded the height, in inches of each player on his team. The results are shown.

marysya [2.9K]

Answer:

3

Step-by-step explanation:

Given:

Team heights (inches):

61, 57, 63, 62, 60, 64, 60, 62, 63

To find: IQRs (interquartile ranges) of the heights for the team

Solution:

A quartile divides the number of terms in the data into four more or less equal parts that is quarters.

For a set of data, a number for which 25% of the data is less than that number is known as the first quartile $(Q_1)$

For a set of data, a number for which 75% of the data is less than that number is known as the third quartile $(Q_3)$

Terms in arranged in ascending order:

$57,60,60,61,62,62,63,63,64$

Number of terms = 9

As number of terms is odd, exclude the middle term that is 62.

$Q_1$ is median of terms $57,60,60,61$

Number of terms (n) = 4

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{60+60}{2}=\frac{120}{2}=60$

So, $Q_1=60$

So, 25% of the heights of a team is less than 60 inches

$Q_3$ is the median of terms $62,63,63,64$

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{63+63}{2}=\frac{126}{2}=63$

So, $Q_3=63$

So, 75% of the heights of a team is less than 63 inches

Interquartile range = $Q_3-Q_1=63-60=3$

The interquartile range is a measure of variability on dividing a data set into quartiles.

The interquartile range is the range of the middle 50% of the terms in the data.

So, 3 is the range of the middle 50% of the heights of the students.

4 0

3 years ago