He will have $840 in 24 months, which is two years.
Answer:
x = -4
Step-by-step explanation:
–2x + 1 – (–7x + 3) = –4 + 3(x – 2)
-2x + 1 + 7x - 3 = -4 + 3x - 6
-2x + 7x - 3x = -4 - 6 - 1 + 3
2x = -8
x = -8/2
x = -4
Answer:
So I am not sure my answers are correct but I hope they help:
A. y = 6x+14
B. 6 people
Answer:
The probability Democrat is selected given that this member favors some type of corporate tax reform is 0.6309.
Step-by-step explanation:
Let us suppose that,
R = Republicans
D = Democrats
I = Independents.
X = a member favors some type of corporate tax reform.
The information provided is:
P (R) = 0.27
P (D) = 0.56
P (I) = 0.17
P (X|R) = 0.34
P (X|D) = 0.41
P (X|I) = 0.25.
Compute the probability that a randomly selected member favors some type of corporate tax reform as follows:
![P(X)=P(X|R)P(R)+P(X|D)P(D)+P(X|I)P(I)\\= (0.34\times0.27)+(0.41\times0.56)+(0.25\times0.17)\\=0.3639](https://tex.z-dn.net/?f=P%28X%29%3DP%28X%7CR%29P%28R%29%2BP%28X%7CD%29P%28D%29%2BP%28X%7CI%29P%28I%29%5C%5C%3D%20%280.34%5Ctimes0.27%29%2B%280.41%5Ctimes0.56%29%2B%280.25%5Ctimes0.17%29%5C%5C%3D0.3639)
The probability that a randomly selected member favors some type of corporate tax reform is P (X) = 0.3639.
Compute the probability Democrat is selected given that this member favors some type of corporate tax reform as follows:
![P(D|X)=\frac{P(X|D)P(D)}{P(X)} =\frac{0.41\times0.56}{0.3639}=0.6309](https://tex.z-dn.net/?f=P%28D%7CX%29%3D%5Cfrac%7BP%28X%7CD%29P%28D%29%7D%7BP%28X%29%7D%20%3D%5Cfrac%7B0.41%5Ctimes0.56%7D%7B0.3639%7D%3D0.6309)
Thus, the probability Democrat is selected given that this member favors some type of corporate tax reform is 0.6309.
<h2>
Answer with explanation:</h2>
Let
be the population mean.
By considering the given information , we have
Null hypothesis : ![H_0: \mu=73.4](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%3D73.4)
Alternative hypothesis : ![H_a: \mu>73.4](https://tex.z-dn.net/?f=H_a%3A%20%5Cmu%3E73.4)
Since alternative hypothesis is right-tailed , so the test is a right-tailed test.
Given : Sample size : n=16 , which is a small sample , so we use t-test.
Sample mean:
;
Standard deviation: ![s=8.4](https://tex.z-dn.net/?f=s%3D8.4)
Test statistic for population mean:
![t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B%5Coverline%7Bx%7D-%5Cmu%7D%7B%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
i.e. ![t=\dfrac{78.3-73.4}{\dfrac{8.4}{\sqrt{16}}}\approx2.333](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B78.3-73.4%7D%7B%5Cdfrac%7B8.4%7D%7B%5Csqrt%7B16%7D%7D%7D%5Capprox2.333)
Using the standard normal distribution table of t , we have
Critical value for
: ![t_{(n-1,\alpha)}=t_{(15,0.01)}=2.602](https://tex.z-dn.net/?f=t_%7B%28n-1%2C%5Calpha%29%7D%3Dt_%7B%2815%2C0.01%29%7D%3D2.602)
Since , the absolute value of t (2.333) is smaller than the critical value of t (2.602) , it means we do not have sufficient evidence to reject the null hypothesis.
Hence, we conclude that we do not have enough evidence to support the claim that answering questions while studying produce significantly higher exam scores.