You have a standard deck of cards. What is the probability of drawing a heart or a face card?

The slope of the line (2,2) and (6,10)

soldi70 [24.7K]

Answer:

2

Step-by-step explanation:

It rose 8 and ran 4 and slope is rise over run.

8/4 = 2

7 0

3 years ago

If x^2-x-6 and x^2+3x-18 hacpve a common factor (x-a) then find the value of a.

DedPeter [7]

Answer:

a = 3

Step-by-step explanation:

Factor both expressions

x² - x - 6

Consider the factors of the constant term (- 6) which sum to give the coefficient of the x- term (- 1)

The factors are - 3 and + 2 , since

- 3 × 2 = - 6 and - 3 + 2 = - 1 , thus

x² - x - 6 = (x - 3)(x + 2)

-----------------------------------

x² + 3x - 18

consider factors of constant term (- 18) which sum to give the coefficient of the x- term (+ 3)

The factors are + 6 and - 3 , since

6 × - 3 = - 18 and 6 - 3 = + 3 , thus

x² + 3x - 18 = (x + 6)(x - 3)

Both expressions have a common factor of (x - 3)

Compare with (x - a ), then a = 3

3 0

2 years ago

Read 2 more answers

For to 30-60-90 triangle below, the length of side AB is 52 cm.

Jlenok [28]

Answer:

Using 60°

Let side CB be x

Cos60= x/52

Cross multiply=

x= 52cos60

x= 26 cm

7 0

3 years ago

Heights of men have a bell-shaped distribution, with a mean of 176 cm and a standard deviation of 7 cm. Using the Empirical Rule

Vaselesa [24]

Answer:

a) 68% of the men fall between 169 cm and 183 cm of height.

b) 95% of the men will fall between 162 cm and 190 cm.

c) It is unusual for a man to be more than 197 cm tall.

Step-by-step explanation:

The 68-95-99.5 empirical rule can be used to solve this problem.

This values correspond to the percentage of data that falls within in a band around the mean with two, four and six standard deviations of width.

a) What is the approximate percentage of men between 169 and 183 cm?

To calculate this in an empirical way, we compare the values of this interval with the mean and the standard deviation and can be seen that this interval is one-standard deviation around the mean:

$\mu-\sigma=176-7=169\\\mu+\sigma=176+7=183$

Empirically, for bell-shaped distributions and approximately normal, it can be said that 68% of the men fall between 169 cm and 183 cm of height.

b) Between which 2 heights would 95% of men fall?

This corresponds to ±2 standard deviations off the mean.

$\mu-2\sigma=176-2*7=162\\\\\mu+2\sigma=176+2*7=190$

95% of the men will fall between 162 cm and 190 cm.

c) Is it unusual for a man to be more than 197 cm tall?

The number of standard deviations of distance from the mean is

$n=(197-176)/7=3$

The percentage that lies outside 3 sigmas is 0.5%, so only 0.25% is expected to be 197 cm.

It can be said that is unusual for a man to be more than 197 cm tall.

3 0

3 years ago

Combine the following fractions. a. 3/5 + 4/15 b. 5/6 - 1/4

Ne4ueva [31]

Hello there! :)

Answer:

1. $\boxed{\frac{13}{15} }$

2. $\boxed{\frac{7}{12} }$

Step-by-step explanation:

1. 3/5+4/15

First, you multiply by numerator and denominator.

3*3=9

5*3=15

$=\frac{9}{15}$

Then, you had to combine by the fractions.

$\frac{9}{15}+\frac{4}{15}$

Add by the numbers from left to right.

9+4=13

13/15 is the final answer.

______________________________________________

2. 5/6-1/4

First, you multiply by 2 both sides of numerator and denominator.

5*2=10

6*2=12

10/12

3/12

$\frac{10}{12}-\frac{3}{12}$

Subtract by the numerator.

10-3=7

7/12 is the final answer.

______________________________

Hope this helps!

Thanks!

Have a nice day! :)

:D

-Charlie

7 0

3 years ago