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timama [110]
3 years ago
15

Please Help Me, Take Your Time.

Mathematics
1 answer:
Doss [256]3 years ago
7 0
<h3>Given</h3>

72 blue, 42 red beads

beads are used to make identical necklaces

<h3>Find</h3>

(a) the greatest number of necklaces that can be made

(b) the number of each color bead in each necklace

<h3>Solution</h3>

You can write and factor the equation

... necklaces = 72 blue + 42 red

... necklaces = 6(12 blue + 7 red)

where 6 is the greatest common factor (GCF) of 72 and 42.

(a) You can make up to 6 identical necklaces. 6 is the largest common factor of 72 and 42. If you were to try to make more, they could not be identical.

(b) Each necklace can consist of 12 blue and 7 red beads. These are the numbers obtained when the total bead count is divided into 6 equal groups.

_____

There are several ways you can find the GCF of two numbers. For small numbers, it is generally feasible to use your knowledge of multiplication tables and factors to choose the largest common factor of two numbers. You can also use Euclid's algorithm, which is to repeatedly compute

... (largest number) modulo (smallest number)

until the result is zero. The final "smallest number" is the GCF.

Here, that looks like

... 72 mod 42 = 30

... 42 mod 30 = 12

... 30 mod 12 = 6

... 12 mod 6 = 0 . . . . . . . so 6 is the GCF

___

Of course, you know that

... 72 = 2³×3²

... 42 = 2×3×7

so, the largest set of common factors is 2×3 = 6.

___

Your graphing calculator may have a function for computing the greatest common divisor (GCD), too. The TI-84 does, for example.

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An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of mov- ing violatio
Rudik [331]

Answer:

a) E(y) = 0.8

b) The average subcharge is $165

Step-by-step explanation:

We are given the following distribution in the question:

y:           0        1        2         3

P(y):   0.50   0.25   0.20   0.05

a) E(y)

E(y) = \displaystyle\sum y_iP(y_i)\\\\E(y) = 0(0.50)+1(0.25)+2(0.20)+3(0.05)\\E(y) = 0.8

b) Expected value of subcharge

Subcharge =

\$110y^2

Expected value of subcharge =

=110E(y^2)\\\\=110\displaystyle\sum y_i^2P(y_i)\\\\=110(0^2(0.50)+1^2(0.25)+2^2(0.20)+3^2(0.05))\\=110(1.5)=165

Thus, the average subcharge is $165

5 0
2 years ago
Anna saves 3 dollars for every 10 dollars she earns.Patrick saves 5 dollars for every 20 dollars he earns.Do the two save money
NNADVOKAT [17]
The two do not save money at the same rate, because if you make common denominators of 20, Anna saves 6 dollars for every 20, while Patrick only saves 5 dollars for every twenty earned. So, Anna ends up saving more than Patrick.
4 0
3 years ago
A random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used. A survey of 500 randomly selecte
vodka [1.7K]

Answer:

We conclude that the mean Ohio score is below the national average.

Step-by-step explanation:

We are given that a random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used.

A survey of 500 randomly selected Ohio scores showed a mean of 20.8. The population standard deviation is 3.

<u><em>Let </em></u>\mu<u><em> = mean Ohio scores.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 21.1      {means that the mean Ohio score is above or equal the national average}

Alternate Hypothesis, H_A : \mu < 21.1      {means that the mean Ohio score is below the national average}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about population standard deviation;

                    T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}}}  ~ N(0,1)

where, \bar X = sample mean Ohio score = 20.8

            \sigma = population standard deviation = 3

            n = sample of Ohio = 500

So, <u><em>test statistics</em></u>  =  \frac{20.8-21.1}{\frac{3}{\sqrt{500}}}  

                              =  -2.24

The value of z test statistics is -2.24.

<em>Now, at 0.1 significance level the z table gives critical value of -1.2816 for left-tailed test.</em><em> Since our test statistics is less than the critical values of z as -2.24 < 1.2816, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the mean Ohio score is below the national average.

4 0
3 years ago
Round 47.219 to the nearest tenth.​
saveliy_v [14]

Answer:

47.2

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
7,014 + 2,415 + 774 + 30 = _____
Cloud [144]

9,459

8,444

23,963

9,304

3,719

7 0
3 years ago
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