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likoan [24]
3 years ago
8

There are total of 84 students in the robotics club and the science club. The science club has 12 more students than the robotic

s club. How many students are in the science club?
Mathematics
1 answer:
kari74 [83]3 years ago
8 0
First off we have x number of students in the robotics club, 12 more in the Science club, and 84 students in total. This means:
Robotics Club = x
Science Club = (x + 12)
Total = 84

Okay, so here's the equation:

x + (x + 12) = 84 ~ Add like terms.
2x + 12 = 84 ~ Subtract 12 by both sides.
2x = 72 ~ Divide both sides by 2.
x = 36 ~ Meaning that the Robotics Club has 36 members.

At this point you can do: 84 - 36 = 48 which gives you the number of members in the Science Club, though...:

36 + (36 + 12) = 84 ~ Add the two in the parenthesis to get your answer.
36 + 48 = 84 ~ 48 is the number of Students in the Science Club.
84 = 84

In conclusion: The Science Club has 48 students.

Hope that helps. ^ ^
{-Ghostgate-}
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A sample of 114 mortgages approved during the current year showed that 36 were issued to a single-earner family or individual. T
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Answer:

a-1) Reject H0 if zcalc > 1.645

a-2) z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561  

a-3) ii. False

b) ii. No

c) iii. n π > 10 and n(1 − π ) > 10

Step-by-step explanation:

1) Data given and notation

n=114 represent the random sample taken

X=36 represent the people that were issued to a single-earner family or individual

\hat p=\frac{36}{114}=0.316 estimated proportion of people that were issued to a single-earner family or individual

p_o=0.28 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

(a-1) H0: π ≤ .28 versus H1: π > .28. Choose the right option. Reject H0 if zcalc > 1.645 Reject H0 if zcalc < 1.645 a b

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.28.:  

Null hypothesis:p\geq 0.28  

Alternative hypothesis:p < 0.28  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

The rejection zone would be on this case :

Reject H0 if zcalc > 1.645

Since is a right tailed test

(a-2) Calculate the test statistic. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.) Test statistic

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561  

(a-3) The null hypothesis should be rejected.

False, since our calculated value is less than our critical value we Fails to reject the null hypothesis

(b) Is this a close decision?

False the calculated value is significantly less than the critical value so we FAIL to reject the null hypothesis with enough confidence.

(c) State any assumptions that are required.

In order to satisfy the conditions we need the following two requirements:

iii. n π > 10 and n(1 − π ) > 10

And are satisfied:

114*0.28=31.92>10

114(1-0.28)=82,08>10

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Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2
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Answer:

(a) f'(x)=-\frac{2}{x^3}

(b) y=-0.25x+0.75

Step-by-step explanation:

The given function is

f(x)=\frac{1}{x^2}                  .... (1)

According to the first principle of the derivative,

f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}

Cancel out common factors.

f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}

By applying limit, we get

f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}

f'(x)=\frac{-2x)}{x^4}

f'(x)=\frac{-2)}{x^3}                         .... (2)

Therefore f'(x)=-\frac{2}{x^3}.

(b)

Put x=2, to find the y-coordinate of point of tangency.

f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}

Substitute x=2 in equation 2.

f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

y-y_1=m(x-x_1)

y-0.25=-0.25(x-2)

y-0.25=-0.25x+0.5

y=-0.25x+0.5+0.25

y=-0.25x+0.75

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0
3 years ago
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