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Oksi-84 [34.3K]
1 year ago
5

Explain how you used the bar model in exercise 2 to solve the problem

Mathematics
1 answer:
lutik1710 [3]1 year ago
3 0

We are given that Grandma has 14 red roses and 7 pink roses.

In order to represent them on bar modal, we need to make a bar with 14 identical sections for 14 red roses.

And for pink roses, we need to make a bar with 7 identical sections for 7 pink roses.

<em>From the bar graph, we can see that bar for red roses have 7 more boxes than bar of pink roses.</em>

<h3>Therefore,  she have 7 more red roses than pink roses.</h3>

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How can patterns be used to determine products of a number and a power of 10?
Sveta_85 [38]
Patterns can be used to determine products of a number of a power of 10 by following it to obtain easily the answer. When you multiply a number by the power of 10 you just have to write the number then add the number of corresponding number of 0s to the end which will be the same number as the power used. 

Hope this helped : )
8 0
2 years ago
Read 2 more answers
Find two numbers that add up to 158. The greater number is 40 more than the lesser number
bija089 [108]

Answer:D

Step-by-step explanation:

6 0
2 years ago
If you are constructing a 95% confidence interval for a normally distributed population when your sample size is 10, what value
Nuetrik [128]

This is something you'll need a T table for, or a calculator that can compute critical T values. Either way, we have n = 10 as our sample size, so df = n-1 = 10-1 = 9 is the degrees of freedom.

If you use a table, look at the row that starts with df = 9. Then look at the column that is labeled "95% confidence"

I show an example below of what I mean.

In that diagram, the row and column mentioned intersect at 2.262 (which is approximate). This value then rounds to 2.26

<h3>Answer:  2.26</h3>

3 0
2 years ago
There are some nickels dimes and quarters in a large piggy bank for every two nickels there are 3 dimes for every two dimes that
creativ13 [48]

We are given

piggy bank has nickels , dimes and quarters

Let's assume

number of nickels =n

every two nickels there are 3 dimes

so, number of dimes are

=\frac{3}{2}n

=1.5n

every two dimes that are 5 quarters there

so, number of quarters are

=\frac{5}{2}\times 1.5n

=3.75n

so, total number of coins = number of nickels + number of dimes +number of quarters

total number of coins =n+1.5n+3.75

there are 500 coins

so, we get

n+1.5n+3.75n=500

now, we can solve for n

6.25n=500

divide both sides by 6.25

so, we get

n=80

number of dimes is 1.5n

=1.5\times 80

=120

number of quarters  is 3.75n

=3.75\times 80

=300

so,

Number of nickels =80

Number of dimes =120

Number of quarters =300............Answer

6 0
2 years ago
Choose the option that best completes the statement below. In finding the number of permutations for a given number of items, __
vodomira [7]
Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have: 3 P_{3}=(3)(2)(1)=6. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.

Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
A
BA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.

So in this case the formula would be \frac{3 P_{3} }{2!}= \frac{(3)(2)(1)}{(2)(1)}= \frac{6}{2}=3. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!


Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.



7 0
2 years ago
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