All categories

2 years ago

From a group of 6 freshmen, 5 sophomores, and 8 seniors, a

Mathematics

2 answers:

ryzh [129]2 years ago

7 0

Answer:

a: 42%

b: 68%

Step-by-step explanation:

Nesterboy [21]2 years ago

4 0

Answer:

1. The probability of selecting a senior is 0.42 (42%).

2. The probability of not selecting a freshman is 0.68 (68%).

Step-by-step explanation:

Total students: 19

Freshmen: 6
Sophomores: 5
Seniors: 8

1. $\frac{8}{19} = 0.42$

2. $\frac{13}{19} = 0.68$

You might be interested in

If a new lacrosse stick is labeled $29, what would the new price be if the sign above it says "10% off"?

olga2289 [7]

The actual price of a new lacrosse stick = $29
Percentage of discount given = 10%
Then
Amount of discount that is given = (10/100) * 29 dollars
   = 29/10 dollars
   = 2.9 dollars
So
The new price of the lacrosse stick = (29 - 2.9) dollars
   = 26.10 dollars
So the new price of the lacrosse stick on display will be $29.10 dollars. I hope there is nothing complicated in the procedure for you to not understand. In future you can always use this procedure for solving similar problems without requiring any additional help from outside.

7 0

4 years ago

Read 2 more answers

Find the midpoint of the line segment given the endpoints. (-1, 10) (-1, 6)

BaLLatris [955]

The midpoint of these two points is (-1, 8).

In order to find the midpoint, you must take the average of each part of the coordinates. You do this by adding them together and dividing by 2. We'll start with the x-coordinates.

(-1 + -1)/2 = -1

Now we'll use the y-coordinates.

(10 + 6)/2 = 8

This gives us the ordered pair of (-1, 8)

8 0

4 years ago

(05.01 MC)

Genrish500 [490]

The answer is attached as a word file

Download doc

5 0

3 years ago

HELP ASAP!! WILL GIVE BRAINLIEST ❤️

mr Goodwill [35]

Answer:

$m=-3$

The slope is negative.

General Formulas and Concepts:

Order of Operations: BPEMDAS
Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (-5, 19)

Point (-1, 7)

Step 2: Find slope m

Substitute: $m=\frac{7-19}{-1-(-5)}$
Simplify: $m=\frac{7-19}{-1+5}$
Subtract/Add: $m=\frac{-12}{4}$
Divide: $m=-3$

8 0

3 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of

Llana [10]

Answer:

Part 1) $csc(\theta)=-\frac{\sqrt{85}}{7}$

Part 2) $sin(\theta)=-\frac{7}{\sqrt{85}}$ or $sin(\theta)=-\frac{7\sqrt{85}}{85}$

Part 3) $tan(\theta)=-\frac{7}{6}$

Part 4) $cos(\theta)=\frac{6}{\sqrt{85}}$ or $cos(\theta)=\frac{6\sqrt{85}}{85}$

Part 5) $sec(\theta)=\frac{\sqrt{85}}{6}$

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

$1+cot^{2}(\theta)=csc^{2}(\theta)$

we have

$cot(\theta)=-\frac{6}{7}$

substitute

$1+(-\frac{6}{7})^{2}=csc^{2}(\theta)$

$1+\frac{36}{49}=csc^{2}(\theta)$

$\frac{85}{49}=csc^{2}(\theta)$ rewrite

$csc(\theta)=-\frac{\sqrt{85}}{7}$ ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

$csc(\theta)=\frac{1}{sin(\theta)}$

we have

$csc(\theta)=-\frac{\sqrt{85}}{7}$

therefore

$sin(\theta)=-\frac{7}{\sqrt{85}}$

$sin(\theta)=-\frac{7\sqrt{85}}{85}$

step 3

Find the value of tan(θ)

we know that

$tan(\theta)=\frac{1}{cot(\theta)}$

we have

$cot(\theta)=-\frac{6}{7}$

therefore

$tan(\theta)=-\frac{7}{6}$

step 4

Find the value of cos(θ)

we know that

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$sin(\theta)=-\frac{7}{\sqrt{85}}$

substitute

$(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1$

$\frac{49}{85}+cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{49}{85}$

$cos^{2}(\theta)=\frac{36}{85}$

$cos(\theta)=\frac{6}{\sqrt{85}}$ ------> the cosine is positive

$cos(\theta)=\frac{6\sqrt{85}}{85}$

step 5

Find the value of sec(θ)

we know that

$sec(\theta)=\frac{1}{cos(\theta)}$

we have

$cos(\theta)=\frac{6}{\sqrt{85}}$

therefore

$sec(\theta)=\frac{\sqrt{85}}{6}$ ----> is positive

8 0

4 years ago