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Oduvanchick [21]
3 years ago
12

) If the pattern shown continues, how many black keys appear on a portable keyboard with 49 white keys? Suggestion: use equivale

nt ratios or a rate table to confirm your answer.

Mathematics
2 answers:
Lorico [155]3 years ago
7 0

In pattern there are 5 black keys and 7 white keys

Given- 49 white keys

49 = 7×7

number of black keys will be equal to = 7×5 = 35 black keys.

andriy [413]3 years ago
5 0

Answer:

35

Step-by-step explanation:

Here we see 5 black keys for every 7 white keys.

So the ratio is 5:7

If we need 49 white keys, find the amount we scale the original ratio by:

    49/7 = 7

So we are scaling by a factor of 7.

The number of black keys would be 5 * the scale of 7. = 35

So there should be 35 black keys.

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A set of numbers is shown below: {0, 0.6, 2, 4, 6} Which of the following shows all the numbers from the set that make the inequ
Lorico [155]
The given equality hold true when x = 2.

Put x = 2 in inequality.

2(2) + 3 = 4+3 = 7 = R.H.S.

For x = 4 and 6, L.H.S(2x+3) is greater than 7.

Hence for x = 2, 4 and 6, the above inequality holds true.

Hope this helps!
3 0
3 years ago
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Seven less than the product of a number and three equal to eleven
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Answer:

7-(3x)=11

Step-by-step explanation:

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A square has a diagonal of length 12. Find the length of each side.
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Answer:

3

Step-by-step explanation:

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7 0
2 years ago
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Step-by-step explanation:

You would measure earth's mass is kg and not in g , mg or mcg

5 0
3 years ago
A poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ag
Katyanochek1 [597]

Answer:

At 5% significance level, it is statistically evident that    there is nodifference in the proportion of college students who consider themselves overweight between the two poll                                  

Step-by-step explanation:

Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.

Let five years ago be group I X and as of now be group II Y

H_0: p_x =p_y\\H_a: p_x \neq p_y

(Two tailed test at 5% level of significance)

                Group I            Group II              combined p

n                  270                 300                         570

favor            120                  140                          260

p                   0.4444            0.4667                   0.4561

Std error   for differene = \sqrt{\frac{0.4561(1-0.4561)}{570} } \\=0.0209

p difference = -0.0223    

Z statistic = p diff/std error =       -1.066

p value =0.2864

Since p value >0.05, we accept null hypothesis.

        At 5% significance level, it is statistically evident that    there is nodifference in the proportion of college students who consider themselves overweight between the two poll                                  

7 0
3 years ago
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