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baherus [9]
2 years ago
15

christine and kaylani are using ribbon to decorate a project in their art class. the ratio of the length of christine's ribbon t

o the length of kaylani's is 7:3. if they have 20 yards of ribbon altogether, how much ribbon does kaylani have?
Mathematics
1 answer:
fenix001 [56]2 years ago
7 0

Answer:

6 yds

Step-by-step explanation:

7 + 3 = 10

christine: 7/10ths of ribbon

0.7(20)=14 yds

kaylani: 3/10ths of ribbon

0.3(20)=6 yds

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Search cos jk and tan jk knowing that sin jk = 8/3
aleksley [76]

Answer:

cosjk = √55 i/3

tanjk = 8/√55 i

Step-by-step explanation:

Given

sin jk = 8/3

According to SOH CAH TOA

Sin theta = opposite/hypotenuse = 8/3

Opposite = 8

hypotenuse = 3

Get the adjacent using the pythagoras theorem

hyp² = opp²+adj²

adj² = hyp² - opp²

adj² = 3² - 8²

adj² = 9-64

adj² = -55

adj = √-55

adj = √55 i (i = √-1)

Get cosjk

cosjk = adj/hyp

cosjk = √55 i/3

Get tanjk

tanjk = opp/adj

tanjk = 8/√55 i

3 0
3 years ago
There are 33 liters of orange juice at a school party. 1010 students want to drink all of the orange juice, and they all want to
skelet666 [1.2K]

Answer: 0.033 liters or 33 ml each

Step-by-step explanation:

There are 33 liters of orange juice and 1,010 students to drink it.

If they are to drink an equal amount, you should divide the quantity of orange juice by the number of students who want to drink it:

= Quantity of orange juice / Number of students drinking

= 33 / 1,010

= 0.033 liters each

In milliliters this would be:

= 0.033 * 1,000 milliliters

= 33 ml each

6 0
3 years ago
Read 2 more answers
Floats in a parade travel down the 5.5 mile long street at a rate of 2.5 miles per hour. How long will it take the floats to com
Rina8888 [55]
It should take 2.2 hours, since 5.5 / 2.5 = 2.2.
6 0
3 years ago
Scores on the test are normally distributed with a mean of 79 and a standard deviation of 8.4. Find the numerical limits for a B
Oksanka [162]

This is not the question, the correct question is:

A humanities professor assigns letter grades on a test according to the following scheme. A: Top 6% of scores B: Scores below the top 6% and above the bottom 59% C: Scores below the top 41% and above the bottom 17% D: Scores below the top 83% and above the bottom 7% F: Bottom 7% of scores.

Scores on the test are normally distributed with a mean of 79 and a standard deviation of 8.4. Find the numerical limits for a B grade. Round your answers to the nearest whole number, if necessary.

Answer:  The numerical limits for grade B is 81 and 92

Step-by-step explanation:

Given that

mean Ж = 79

Standard deviation S = 8.4

B: Scores below the top 6% and above the bottom 59%

To find the numerical value for Grade B

Below the top bottom 6% ( 0.06) is

p ( X < Ж ) = 1 - 0.06

p ( X < Ж ) = 0.94

therefore

P( (X - ц)/S < (X - Ж) / S)) = 0.94

(X - 79) / 8.4 = 1.5548 ( from normal distribution table )

X - 79 = 13.0603

X = 13.0603 + 79

X = 92.06 ≈ 92 ( nearest whole number)

Above bottom 59% ( 0.59) is

p ( X < Ж ) = 0.59

therefore

P( (X - ц)/S < (X - Ж) / S)) = 0.59

(X - 79) / 8.4 = 0.2275 ( from normal distribution table )

X - 79 = 1.9114

X = 1.9114 + 79

X = 80.91 ≈ 81 ( nearest whole number)

So the numerical limits for grade B is 81 and 92

3 0
3 years ago
Can someone please help me with 2 and explain the steps to 3 thanks!
Gelneren [198K]

Answer:

#2 is a ratio so you gotta add.

#3 it’s just adding it for example 2p + 5a = 7pa

7 0
3 years ago
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