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klemol [59]
3 years ago
13

If x varies directly with y and y is 10 when x is 8, what is x when y is 18?

Mathematics
2 answers:
swat323 years ago
5 0
Knowing that x varies directly with y, we know the basic skeleton of the equation is y = mx. 

To find m, plug in the known coordinate (8,10), where the x-coordinate is 8 and the y-coordinate is 10.
y=mx \\ 10=m8 \\ (10)/8=(8m)/8 \\ m= \frac{10}{8} = \frac{5}{4}

Then, plug in m to the equation to get y = (5/4)x.

Finally, plug in the y-coordinate 18 to the equation and solve for x!
y= \frac{5}{4}x \\ 18= \frac{5}{4}x \\  \frac{4}{5}(18)=( \frac{5}{4}x) \frac{4}{5} \\ x= \frac{4*18}{5}  \\  x= \frac{72}{5}

Your exact answer is x = 72/5. However, if your teacher prefers decimals, then x = 14.4

Hope this helps!
Vesnalui [34]3 years ago
3 0

y varies directly with x
y=kx
10=k(15)
k=10/15
y=(10/15)x
when x=6
y=(10/15)(6)=60/15=4 
 x varies inversely as y
x=k/y
2=k/8
k=16
x=16/y
when y=17

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Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

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Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men      Women

The sample mean : 24.51      22.69

Sample standard deviation : 4.48    3.86

Sample size : 35    40

<em>Let </em>\mu_1<em> = mean number of times men order take-out dinners in a month.</em>

<em />\mu_2<em> = mean number of times women order take-out dinners in a month</em>

(a) So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0     {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, H_A : \mu_1-\mu_2\neq 0     {means that there is difference in the mean number of times men and women order take-out dinners in a month}

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where, \bar X_1 = sample mean for men = 24.51

\bar X_2 = sample mean for women = 22.69

s_1 = sample standard deviation for men = 4.48

s_2 = sample standard deviation for women = 3.86

n_1 = sample of men = 35

n_2 = sample of women = 40

Also,  s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }  =  \sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2}  }{35+40-2} } = 4.16

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                     P-value = P( t_7_3 > 1.89) = 0.0331

So, P-value for two tailed test is = 2 \times 0.0331 = <u>0.0662</u>

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