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seraphim [82]
2 years ago
5

The quantities xxx and yyy are proportional. xxx yyy 5.85.85, point, 8 5.85.85, point, 8 7.57.57, point, 5 7.57.57, point, 5 11.

211.211, point, 2 11.211.211, point, 2 Find the constant of proportionality (r)(r)left parenthesis, r, right parenthesis in the equation y=rxy=rxy, equals, r, x. r =r=r, equals
Mathematics
2 answers:
Paul [167]2 years ago
5 0

Question:

The quantities x and y are proportional.

x y

5.8 7.5

11.2

Find the constant of proportionality (r) in the equation y=rx

Answer:

The constant of proportionality is 75/58 or 1.29

Step-by-step explanation:

Given

The table above

Required

Find the constant of proportionality

The question has an incomplete table but it can still be solved because x and y are proportional.

Given that

y = rx

Make r the subject of formula

Divide through by x

y/x = rx

y/x = r

r = y/x

When y = 7.5, x = 5.8

Substitute these values

r = y/x becomes

r = 7.5/5.8

Multiply denominator and numerator by 10

r = (7.5 * 10)/(5.8 * 10)

r = 75/58

In this case, it's best to leave the answer in fraction.

However, it can be solved further.

r = 75/58

r = 1.29 (Approximated)

Hence, the constant of proportionality is 75/58 or 1.29

QveST [7]2 years ago
5 0

Answer:

the correct answer is 1

hope this helps :D

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Answer:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

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Step-by-step explanation:

We have the following info given:

\bar X_1 = 240 sample mean for medium Pizzas from Prim's

\bar X_2 = 210 sample mean for medium Pizzas from Pizza Place

s_1 =8.6 sample deviation for Prim's

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The confidence interval for the difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

And for the 95% confidence we need a significance level of \alpha=1-0.95=0.05 and \alpha/2 =0.025, the degrees of freedom are given by:

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t_{\alpha/2}= 1.984

And replacing we got:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

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