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Citrus2011 [14]
3 years ago
6

Find the value of y for a given value of x, if y varies directly with x. If y = 2.52 when x = 8.4, what is y when x = 2.7?  (1 p

oint) 9 –9 0.81 –0.81
Mathematics
2 answers:
Levart [38]3 years ago
5 0

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form (y/x)=k\ or\ y=kx

in this problem we have

y = 2.52\ when\ x = 8.4

so

<u>Find the value of the constant k</u>

(y/x)=k

substitute

k=(2.52/8.4)

k=0.3

the equation is equal to

y=0.3x

<u>Find the value of y when x=2.7</u>

substitute the value of x in the equation

y=0.3*2.7

y=0.81

therefore

<u>the answer is</u>

0.81


Gwar [14]3 years ago
4 0
 <span>8.4/2.52 = 2.7/y 
cross multiply: 
8.4y = 6.80 
y = 6.80/8.4 =  choice (D) </span>
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A cookie factory monitored the number of broken cookies per pack yesterday.
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Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given  

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20  

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } } \leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

8 0
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