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Alona [7]
3 years ago
13

What is the slope of the line that contains the points (-2, 7) and (2, 3)? O A. -4 OB. 4 O C. -1 O D. 1 SUBM​

Mathematics
2 answers:
mash [69]3 years ago
7 0

Answer:

slope = (y2-y1)/(x2-x1) using your ordered pairs this would be (3 - 7)/(2- -2) so you would get (-4) / 4 which equals -1

Step-by-step explanation:

viktelen [127]3 years ago
4 0

Answer:

- 1

Step-by-step explanation:

slope =  \frac{3 - 7}{2 - ( - 2)}  =  \frac{ - 4}{2 + 2}  =   - \frac{4}{4}  =  - 1 \\

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Unit 3 | Lesson 9
Charra [1.4K]
The formula d = rt gives the distance traveled in time t at rate r. A bicyclist rides for 1.5 hours and travels 25 miles. 
d= D. 37.5 miles per hour
r= 25
t=1.5
25*1.5
=37.5

D.
37.5 miles per hour
5 0
3 years ago
A random sample of 850 births included 434 boys. Use a 0.10 significance level to test the claim that 51.5​% of newborn babies a
Molodets [167]

Answer:

A

Null hypothesis: H0: p = 0.515

Alternative hypothesis: Ha ≠ 0.515

z = -0.257

P value = P(Z<-0.257) = 0.797

Decision; we fail to reject the null hypothesis. That is, the results support the belief that 51.5​% of newborn babies are​ boys

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 90% confidence interval) ---- reject Null hypothesis

Z score < Z(at 90% confidence interval) ------ accept Null hypothesis

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

For the case above;

Null hypothesis: H0: p = 0.515

Alternative hypothesis: Ha ≠ 0.515

Given;

n=850 represent the random sample taken

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 850

po = Null hypothesized value = 0.515

p^ = Observed proportion = 434/850 = 0.5106

Substituting the values we have

z = (0.5106-0.515)/√(0.515(1-0.515)/850)

z = −0.256677

z = −0.257

To determine the p value (test statistic) at 0.10 significance level, using a two tailed hypothesis.

P value = P(Z<-0.257) = 0.797

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = -0.257) which falls with the region bounded by Z at 0.10 significance level. And also the one-tailed hypothesis P-value is 0.797 which is greater than 0.10. Then we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that at 10% significance level the null hypothesis is valid.

5 0
3 years ago
What is the solution to the equation 1.6m − 4.8 = −1.6m?
Alla [95]
<span>1.6m − 4.8 = −1.6m
<u>- 1.6m           -1.6m
</u>        -4.8  =  -3.2m
<u>
</u>divide both sides by -3.2

m = 1.5</span>
7 0
3 years ago
The quantities x and y are proportional. Find the constant of proportionality r in the equation y = rx
ELEN [110]

Answer:

r = 7

Step-by-step explanation:

To solve this, we can plug in a pair of x and y values and solve for r.

y = rx | Plug in a pair

42 = r*6 | Now divide both sides by 6

7 = r.

We can test this by plugging in r with a pair.

y = (7)x

77 = 7*11, 77 = 77, This equation is correct.

7 0
3 years ago
Read 2 more answers
Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
pantera1 [17]

This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k

Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

5 0
4 years ago
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