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Dimas [21]
3 years ago
12

Variable y varies directly with variable x, and y = 6 when x = 9.

Mathematics
2 answers:
lesantik [10]3 years ago
5 0
X and y are proportional hence when y is multiplied by a number, x is multiplied by the same number.

18=6*3 hence when y=18,x=9*3=27

x=27
kap26 [50]3 years ago
4 0
So you can set up a proportion like
x/y=x/y
9/6=x/18
simplify 9/6=3/2
3/2=x/18
multiply both side by 18
(3x18)/2=x
54/2=x
x=27
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19. Rick shared his bag of grapes withfriends. He gave /210of the bag to Melissaand /410 of the bag to Ryan. What fractionof the
kvv77 [185]

Rick started with 10/10 of the grapes because he had all of them at the start.

Then he gave 2/10 and 4/10 of the grapes away.

10/10 - 2/10 - 4/10 = 4/10 = 2/5 left

answer: Rick had 2/5 of the grapes left.

4 0
3 years ago
Two minor league baseball players got a total of 210 hits. Washington had 12 more hits than Sanchez. find the number of hits fro
mote1985 [20]
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8 0
3 years ago
What value, written as a decimal, should lena use as the common ratio? 1.0 1.5 2.5 2.0
mariarad [96]

The Lena has no common ratio for the sequence 1.0 1.5 2.5 2.0

According to the question,

1.0  1.5  2.5  2.0

a = 1.0

r = 1.5/1 = 1.5

r = 2.5/1.5 = 1.6667

r = 2/2.5 = 0.8

common ratio is different, so given sequence is not geometric sequence.

Hence,  the Lena has no common ratio for the sequence 1.0 1.5 2.5 2.0.

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6 0
2 years ago
A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit surv
lys-0071 [83]

Answer:

Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

Step-by-step explanation:

Given - A university researcher wants to estimate the mean number

            of  novels that seniors read during their time in college. An exit

            survey was conducted with a random sample of 9 seniors. The

            sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

              been met, which of the following is a 94.645% confidence

              interval for the population mean number of novels read by

              all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

                                = n - 1

                                = 9 - 1

                                = 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

   =1 - 0.94645

  =0.05355 ≈ 0.05

⇒α = 0.5

Now,

\frac{\alpha}{2} = \frac{0.05}{2}

  = 0.025

Then,

t_{\frac{\alpha}{2}, df }  = 2.306

∴ we get

Population mean = x⁻ ± t_{\frac{\alpha}{2}, df } ×\frac{s}{\sqrt{n} }

                           = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

⇒Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

3 0
3 years ago
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Suppose there are 17 jelly beans in a box-2 red, 6 blue, 4 white, and 5 green.
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