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Anna11 [10]
3 years ago
11

a spool of ribbon holds 6.75 meters a craft club buys 21 spools if the club uses 76.54 meters to complete a project how much rib

bon will be left
Mathematics
1 answer:
konstantin123 [22]3 years ago
5 0
Well, you would need to multiply 6.75 and 21, since there are 21 spools, you then simply subtract the spool used for the project and the is your answer, and don't forget to label
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Gavyn and Alex win some money and share it in the ratio 3:2. Gavyn gets £27. How much did they win in total?
velikii [3]
27÷3=9
9 is the original number.
9×2=18

the answer is 18.
if you wanted to check your work, 27:18 simplifies down to 3:2.
7 0
4 years ago
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Assume that blood pressure readings are normally distributed with a mean of 124 and a standard deviation of 6.4. If 64 people ar
Murrr4er [49]

Answer:

99.38%

Step-by-step explanation:

We have that the mean (m) is equal to 124, the standard deviation (sd) 6.4 and the sample size (n) = 64

They ask us for P (x <126)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / (sd / (n ^ 1/2))

We have all these values, replacing we have:

z = (126 - 124) / (6.4 / (64 ^ 1/2))

z = 2.5

With the normal distribution table (attached), we have that at that value, the probability is:

P (z <2.5) = 0.9938

The probability is 99.38%

4 0
3 years ago
Three times a number, minus 5, is equal to two times the number, plus 4. Find the number
fiasKO [112]
I believe the answer is 14. This is a hard one.
4 0
3 years ago
A 0. 0427 kg racquetball is moving 22. 3 m/s when it strikes a stationary box. The ball bounces back at 11. 5 m/s, while the box
andreyandreev [35.5K]

To solve the question we must know about the conservation of momentum.

<h2>Conservation of Momentum</h2>

According to the law of conservation of momentum, when two bodies collide with each other then their momentum is conserved if no external force acts on the system. therefore, the combined momentum of the two objects before collision will be equal to the combined momentum of the two objects after the collision.

the momentum of the two objects before the collision

                       = momentum of the two objects after the collision

The mass of the box is 0.30141 kg.

<h2>Explanation</h2>

Given to us

  • Mass of racquetball, m = 0.0427 kg,
  • Velocity of racquetball before Collision,  \bold{u_1} = 22.3 m/s,
  • Velocity of box before Collision,\bold{ u_2}  = 0 m/s,
  • Velocity of racquetball after Collision, \bold{v_1} = 11.5 m/s,
  • Velocity of box after Collision, \bold{v_2} = 1.53 m/s,

<h3>Assumption</h3>

Let the mass of the box be M.

<h3>Conservation Of Momentum</h3>

Momentum before collision = Momentum after collision

mu_1 + Mu_2 = mv_1 + Mv_2

given, the box was stationary before the collision,

mu_1 + Mu_2 = mv_1 + Mv_2\\\\&#10;mu_1 + 0 = mv_1 + Mv_2\\\\&#10;mu_1  - mv_1 = Mv_2\\\\&#10;M= \dfrac{mu_1  - mv_1}{v_2}

Substituting the values,

M= \dfrac{mu_1  - mv_1}{v_2}\\\\&#10;M = \dfrac{(0.0427 \times 22.3)  - (0.0427\times 11.5)}{1.53}\\\\&#10;M = 0.30141\ kg

Hence, the mass of the box is 0.30141 kg.

Learn more about Conservation of Momentum:

brainly.com/question/2141713

3 0
3 years ago
I NEED HELP first awnser gets brainliest
Levart [38]

Answer:

uhhhhh B i think

Step-by-step explanation:

i did it

7 0
3 years ago
Read 2 more answers
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