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Elis [28]
3 years ago
9

You are making necklaces for your friends. You have 72 blue beads and 42 red beads. Each friend will receive an identical neckla

ce, and all beads must be used.
(A) What is the greatest number of friends who can receive a necklace? Explain your reasoning.
(B) How many of each color bead will each friend receive on the necklace? Explain your reasoning.

Help me for free minecoins T-T
Mathematics
1 answer:
Lilit [14]3 years ago
8 0

Answer:

A) 6 friends

B) 12 blue beads and 7 red beads

Step-by-step explanation:

<em>Step 1: Find the total number of beads</em>

Blue beads + red beads = Total beads

72 + 42 = 114

<em>Step 2: Find the ratio of blue beads to red beads</em>

Blue/red = 72/42 = 12/7

<em>Step 3: Find the total number of beads in each necklace</em>

Add the ratio found in step 2

12 + 7 = 19 beads in each necklace

<em>Step 4: Find the number of necklaces</em>

Total beads/beads in each necklace

114/19 = 6

There will be 6 identical necklaces.

A) 6 friends will receive a necklace

B) Each friend will receive 12 blue beads and 7 red beads on each necklace.

!!

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Answer:

63.08 km

Step-by-step explanation:

To find how far you drive, add up the km from only Monday and Tuesday, since you aren't driving on Friday

34.85 + 28.23

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So, you would drive 63.08 km

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3 years ago
We wish to give a 90% confidence interval for the mean value of a normally distributed random variable. We obtain a simple rando
coldgirl [10]

Answer: (9.27025,\ 11.12975)

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean : \overline{x}=10.2

sample standard deviation : s= 1.5

Significance level ; \alpha= 1-0.90=0.10

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value = t_{\alpha/2, df}=t_{0.05 , 8}=1.8595

Confidence interval for the population mean :

\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}

90% confidence interval for the mean value will be :

10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}

10.2\pm (1.8595)\dfrac{1.5}{3}

10.2\pm (1.8595)(0.5)

10.2\pm (0.92975)

(10.2-0.92975,\ 10.2+0.92975)

(9.27025,\ 11.12975)

Hence, the 90% confidence interval for the mean value= (9.27025,\ 11.12975)

6 0
3 years ago
What is 2(x-3) = 14<br> X= ?
HACTEHA [7]

Answer:

X=4 hope this helped :)

Step-by-step explanation:

2(x-3)=14

2x-6=14

2x= 14-6

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erastova [34]

Answer:

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Step-by-step explanation:

4 7/10 + 6/8

Simplify the fractions by dividing the second top and bottom  by 2

4 7/10 + 3/4

Getting a common denominator of 20

4 7/10 *2/2  + 3/4 * 5/5

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4 29/20

4 + 20/20 + 9/20

4+1+9/20

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3 years ago
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Assume the mean life of an iPhone is 5 years (60 months)and a standard deviation of 8 months If the age in years at which an iPh
melamori03 [73]

Answer:

Answer:

safe speed for the larger radius track u= √2 v

Explanation:

The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.

Also given that r_1= smaller radius

r_2= larger radius curve

r_2= 2r_1..............i

let u be the speed of larger radius curve

now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=

r

1

mv

2

=

r

2

mu

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................ii

form i and ii we can write

v^2= \frac{1}{2} u^2v

2

=

2

1

u

2

⇒u= √2 v

therefore, safe speed for the larger radius track u= √2 v

8 0
2 years ago
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