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2 years ago

Using the distributive property, what would be the expression (I DONT NEED ANSWER) 1/4(8 + x + 4).

Mathematics

2 answers:

xxMikexx [17]2 years ago

5 0

Answer:x/4+3

Step-by-step explanation:

8/1×1/4=8/4=2

X/1×1/4=x/4

4/1×1/4=1

Licemer1 [7]2 years ago

5 0

Answer:

2 + $\frac{1}{4} x$ + 1

Step-by-step explanation:

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I need the answer for this problem please

Savatey [412]

Answer:

Step-by-step explanation:

Find all probabilities:

A. False

$Pr(\text{red shirt}|\text{large shirt})=\dfrac{\text{number red large shirts}}{\text{number large shirts}}=\dfrac{42}{77}=\dfrac{6}{11}\\ \\Pr(\text{large shirt})=\dfrac{\text{number large shirts}}{\text{number shirts}}=\dfrac{77}{165}=\dfrac{7}{15}$

B. True

$Pr(\text{blue shirt}|\text{large shirt})=\dfrac{\text{number blue large shirts}}{\text{number large shirts}}=\dfrac{35}{77}=\dfrac{5}{11}\\ \\Pr(\text{blue shirt})=\dfrac{\text{number blue shirts}}{\text{number shirts}}=\dfrac{75}{165}=\dfrac{5}{11}$

C. False

$Pr(\text{shirt is medium and blue})=\dfrac{\text{number medium and blue shirts}}{\text{number shirts}}=\dfrac{48}{165}=\dfrac{16}{55}\\ \\Pr(\text{medium shirt})=\dfrac{\text{number medium shirts}}{\text{number shirts}}=\dfrac{88}{165}=\dfrac{8}{15}$

D. False

$Pr(\text{large shirt}|\text{red shirt})=\dfrac{\text{number red large shirts}}{\text{number red shirts}}=\dfrac{42}{90}=\dfrac{7}{15}\\ \\Pr(\text{red shirt})=\dfrac{\text{number red shirts}}{\text{number shirts}}=\dfrac{90}{165}=\dfrac{6}{11}$

4 0

3 years ago

What are the 1st and 3rd quartiles?<br> what is the interquartile range?

FrozenT [24]

Answer:

The first quartile is 5

The third is 8

The interquartile is the difference between the first and third quartile.

The interquartile is 3

Step-by-step explanation:

First, make a list of all the numbers. To find the first quartile, you have to first find the median. The median is the first 7 in the list.

Everything before 7 is the lower percentile of the range.

The firsr quartile is the median of the lower percentile.

The lower list is:

2,3,4,4,5,6,6,6

The median of this list is 5.

The process for finding the third quartile is the same except the list is the higher percentile which is the list of numbers above the median.

7 0

2 years ago

Read 2 more answers

GCF And LCM of 24, 36, 45

faltersainse [42]

Answer:

Step-by-step explanation:

First lets prime factorize each number

24 = 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

45 = 3 x 3 x 5

GCF = 3 (3 is the only common prime number factor in the 3 numbers)

LCM = 2 x 2 x 2 x 3 x 3 x 5

= 24 x 3 x 5

= 72 x 5

= 360

Happy to help :)

4 0

2 years ago

A taxi fare has a fixed charge of $1.75 and an additional charge $0.45 per mile. A fare for m miles is $13.90. Which equation co

Anestetic [448]

Answer:

B) 1.75 + 0.45m = 13.90

Step-by-step explanation:

To calculate the Taxi fare, we need to sum the fixed charge and the additional charge. Fixed charge ever is $1.75. The additional charge changes with the miles traveled (one mile costs $0.45, two miles $0.90, and so on), so we need multiply the charge ($0.45) by the miles m traveled.

1.75 + 0.45 m

The problem says that the fare was equals to $13.90 so the final equation is

1.75 + 0.45m = 13.90

6 0

3 years ago

drag the gray arrow so that it is pointing midway between horizontal and vertical, heading downward and toward the right. extend

alex41 [277]

The speed of the star as compared to the speed of the star relative to earth is the frequency of light emitted by the source.

A star moving away from the observer (say on Earth) will have its emitted light red-shifted (move to lower frequency), i.e., the spectrum moving to left if we consider the intensity of light is plotted as a function of frequency with the frequency increasing from left to right.

A star moving towards the observer (say on Earth) will have its emitted light blue-shifted (move to higher frequency), i.e., the spectrum moving to the right if we consider the intensity of light is plotted as a function of frequency with the frequency increasing from left to right.

A star moving perpendicular to the observer (say on Earth) will have its emitted light will no change with respect to the spectrum of the stationary star.

These are due to the Doppler effect, the equation of frequency detected by the observer as

$f_o = f_s\sqrt{\frac{1-\beta }{1+\beta } }$ where $\beta = \frac{v}{c}$ , v the relative velocity of the moving source (the star here), f^s is the frequency of light emitted by the source.

Hence the answer is the speed of the star as compared to the speed of the star relative to earth is the frequency of light emitted by the source.

To learn more about frequency click here brainly.com/question/26177128

#SPJ4

7 0

10 months ago