Since the price is going down, it's going to be a percent decrease. Now to find the percent, we divide the "after" price by the "before" price.

240 / 320 = 0.75 = 75%

The item had a 75% decrease.

3 0

3 years ago

1. Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than

zaharov [31]

1. C(x, y) = (7.3, –3.9)

2. C(x, y) = (17, –1.5)

Solution:

Question 1:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction $\frac{3}{8}$ .

Point divides segment in the ratio formula:

$$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)$

Here, $x_1=3, y_1=-5, x_2=19, y_2=-1$ and m = 3, n = 8

$$C(x, y)=\left(\frac{3\times19+8\times3}{3+8} , \frac{3\times(-1)+8\times(-5)}{3+8}\right)$

$$=\left(\frac{57+24}{11} , \frac{-3-40}{11}\right)$

$$=\left(\frac{81}{11} , \frac{-43}{11}\right)$

C(x, y) = (7.3, –3.9)

Question 2:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction $\frac{3}{8}$ .

Point divides segment in the ratio formula:

$$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)$

Here, $x_1=3, y_1=-5, x_2=19, y_2=-1$ and m = 7, n = 1

$$C(x, y)=\left(\frac{7\times19+1\times3}{7+1} , \frac{7\times(-1)+1\times(-5)}{7+1}\right)$

$$=\left(\frac{133+3}{8} , \frac{-7-5}{8}\right)$

$$=\left(\frac{136}{8} , \frac{-12}{8}\right)$

C(x, y) = (17, –1.5)

8 0

3 years ago

A multiple-choice examination has 20 questions, each with five possible answers, only one of which is correct. Suppose that one

kari74 [83]

Answer:

The probability that the student answers at least seventeen questions correctly is $8.03\times 10^{-10}$ .

Step-by-step explanation:

Let the random variable X represent the number of correctly answered questions.

It is provided all the questions have five options with only one correct option.

Then the probability of selecting the correct option is,

$P(X)=p=\frac{1}{5}=0.20$

There are n = 20 question in the exam.

It is also provided that a student taking the examination answers each of the questions with an independent random guess.

Then the random variable can be modeled by the Binomial distribution with parameters n = 20 and p = 0.20.

The probability mass function of X is:

$P(X=x)={20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x};\ x =0,1,2,3...$

Compute the probability that the student answers at least seventeen questions correctly as follows:

$P(X\geq 17)=P (X=17)+P (X=18)+P (X=19)+P (X=20)$

$=\sum\limits^{20}_{x=17}{{20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x}}\\\\=0.00000000077+0.000000000032+0.00000000000084+0.000000000000042\\\\=0.000000000802882\\\\=8.03\times10^{-10}$

Thus, the probability that the student answers at least seventeen questions correctly is $8.03\times 10^{-10}$ .

4 0

3 years ago