All categories

3 years ago

Write a fraction that is equivalent to a terminating decimal between 0.5 and 0.75

1 answer:

fomenos3 years ago

7 0

0.5 = 1/2 = 2/4 = 4/8 = 8/16 = 16/320.75 = 3/4 = 6/8 = 12/16 = 24/32
Then the fractions in between are 5/8, 9/16, 11/16, 17/32, 19/32, 21/32, 23/32

You might be interested in

PLEASE HELP!!!

iogann1982 [59]

Her new car costed $74,855.96

Step-by-step explanation:

Sara just bought a new car

This car cost $2,362.12 less than twice the price of her first car
She paid 38,609.04 dollars for her first car

we need to find how much her new car costed

∵ The cost of her first car = $38,609.04

∵ The cost of the new car is $2,362.12 less than twice the

cost of her first car

- That means multiply the cost of the 1st car by 2 and then

subtract 2,362.12 from the product

∴ The cost of the new car = 2(38,609.04) - 2,362.12

∴ The cost of the new car = 77218.08 - 2,362.12

∴ The cost of the new car = 74,855.96

Her new car costed $74,855.96

Learn more:

You can learn more about the word problems in brainly.com/question/10557938

#LearnwithBrainly

7 0

3 years ago

Personnel selection. Suppose that 6 female and 6 male applicants have been successfully screened for 5 positions. If the 5 posit

iragen [17]

Answer:

The answer is below

Step-by-step explanation:

The possible ways of choosing r items from a total of n item is given as C(n,r) = $\frac{n!}{(n-r)!r!}$

Suppose that 6 female and 6 male applicants have been successfully screened for 5 positions.

Since they are 12 finalists and only 5 positions, the total number of possible outcomes is:

$Total\ number\ of \ outcome (n(s))=C(12,5)=\frac{12!}{(12-5)!5!} =792$

A) 3 females and 2 males

The possible ways of choosing 3 females and 2 males is given as:

$n(e)=C(6,3)*C(6,2) = \frac{6!}{(6-3)!3!}* \frac{6!}{(6-2)!2!}=20*15=300$

probability of selecting 3 females and 2 males P(E) = $\frac{n(e)}{n(s)}=\frac{300}{792}=0.379$

B) 4 females and 1 males

The possible ways of choosing 4 females and 1 males is given as:

$n(e)=C(6,4)*C(6,1) = \frac{6!}{(6-4)!4!}* \frac{6!}{(6-1)!1!}=15*6=90$

probability of selecting 4 females and 1 males P(E) = $\frac{n(e)}{n(s)}=\frac{90}{792}=0.1136$

C) 5 females

The possible ways of choosing 5 females is given as:

$n(f)=C(6,5) = \frac{6!}{(6-5)!5!}=6$

probability of selecting 5 females = $\frac{n(f)}{n(s)}=\frac{6}{792}$

(D) At least 4 females

This can be done by choosing 4 females out of 6 females i.e. C(6,4) , 5 females out of 6 females i.e C(6,5) and 1 male out of six male i.e (6,1)

The possible ways of choosing atleast 4 females is given as:

$n(h)=C(6,5)*C(6,4)*C(6,1)= \frac{6!}{(6-5)!5!}* \frac{6!}{(6-4)!4!}* \frac{6!}{(6-1)!1!}=6*15*6=540$

probability of selecting atleast 4 females = $\frac{n(h)}{n(s)}=\frac{540}{792}$

5 0

3 years ago

Simplify given

Ksju [112]

Step-by-step explanation:

Hey there!

Please see attached picture for your answer.

Hope it helps!

6 0

3 years ago

Three sixth-grade students competed in the 100-meter dash. The table shows their times.

Sergeeva-Olga [200]

Answer:

It is 12.9, 13.1, 13.2 C.

Step-by-step explanation:

5 0

3 years ago

Given: AB ∥ DC

RSB [31]

Answer:

$A=1,720.16\ units^2$

Step-by-step explanation:

we know that

The area of the trapezoid is equal to

$A=\frac{1}{2}(DC+AB)DE$

step 1

Find the measure of angle DAE

m∠ADC+m∠DAE=180° -----> by consecutive interior angles

we have

m∠ADC = 134°

substitute

134°+m∠DAE=180°

m∠DAE=180°-134°=46°

step 2

In the right triangle ADE

Find the length side AE

cos(∠DAE)=AE/AD

$AE=cos(46\°)(40)\\AE=27.79\ units$

step 3

In the right triangle ADE

Find the length side DE

sin(∠DAE)=DE/AD

$DE=sin(46\°)(40)\\DE=28.77\ units$

step 4

Find the area of ABCD

$A=\frac{1}{2}(DC+AB)DE$

we have

$DC=32\ units\\AB=DC+2(AE)=32+2(27.79)=87.58\ units\\DE=28.77\ units$

substitute

$A=\frac{1}{2}(32+87.58)28.77$

$A=1,720.16\ units^2$

5 0

4 years ago