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Lynna [10]
3 years ago
10

The line that goes through the point (5,-2) and (8,4)

Mathematics
1 answer:
scZoUnD [109]3 years ago
5 0
So for this equation you will want to graph the points (5,-2) and (8,4). I have taken the liberty of doing the graph for you.

With this I have put the two dots on the 2 locations they wanted. So then what you need to do is draw a straight line from dot one to dot two.

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(3-22)-6 + [28-(4)21-7l÷6+o A. 32C > B. 23OC, 55D. 46ー 23563254S ABCD
anzhelika [568]
I got b
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3 0
3 years ago
A car covers a distance of 450 m in 1 minute whereas a train covers
velikii [3]
Speed = distance/ time
Car : 450m/1min
Train: 69km/45 min
7 0
3 years ago
If x varies inversely as y 2, and x = 2 when y = 20, find x when y = 5.
Dominik [7]

Step-by-step explanation:

x =  \frac{k}{y}

x varies inversely as y. This means that x is equal to the inverse of y which can be translated as x = 1/y but the problem said that x is <u>n</u><u>o</u><u>t</u><u> </u><u>e</u><u>x</u><u>a</u><u>c</u><u>t</u><u>l</u><u>y</u><u> </u><u>e</u><u>q</u><u>u</u><u>a</u><u>l</u><u> </u>to the inverse of y. Therefore, we need to introduce a factor k. Factor k will represent the word "varies" in the problem.

To illustrate:

x =  \frac{k}{y}

Since we don't have the value of k, we have to solve for it by using the first values of x and y which was given as x=2 and y=20.

x =  \frac{k}{y}  \\ 2 =  \frac{k}{20}  \\ k = 2(20) = 40

Now that we have the value of k, we can now solve for the value of x when y=5.

x =  \frac{k}{y}  \\  =  \frac{40}{5}  \\  = 8

Therefore, when y is 5, the value of x is 8.

6 0
3 years ago
Condense the expression into a
liberstina [14]

2(\log 18-\log 3)+\dfrac 12 \log \dfrac{1}{16}=2 \log \dfrac{18}{3}+\log \sqrt{\dfrac{1}{16}}\\=2 \log 6+\log \dfrac{1}{4}=\log 6^2+\log \dfrac{1}{4}=\log 36+\log \dfrac{1}{4}\\=\log \dfrac{36}{4}=\log 9

7 0
1 year ago
Researchers studying the effect of diet on growth would like to know if a vegetarian diet affects the height of a child. Twelve
Eva8 [605]

Answer:

a)  t_{0.035},11=\pm2.201

b)  t=-2.963

c)  Reject\ H_0\ when\ \alpha=0.05

d)   t

   -2.963

Step-by-step explanation:

From the question we are told that:

Sample size n=12

Mean \=x=42.5

Standard deviation \sigma=3.8

Population mean \mu=45.75

Significance \alpha=0.05

 

Generally the hypothesis given by

H_0;\mu=45.75\\H_1:\neq =45.75

Generally the equation for test statistics is mathematically given by

t=\frac{\=x-\mu}{\sigma/\sqrt{n} }

t=\frac{42.5-45.75}{3.8/\sqrt{12} }

t=-2.963

Generally the Critical value is mathematically given by

t_{\alpha/2},d_t

\alpha=0.05 \\\alpha/2=0.025\\d_t=n-1=11

t_{0.035},11

From table

t_{0.035},11=\pm2.201

Therefore

t

-2.963

Reject\ H_0\ when\ \alpha=0.05

4 0
3 years ago
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