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2 years ago

Plzz answer these!!!!!!!!! will give brainliest!!!!!!!!

Mathematics

1 answer:

olga55 [171]2 years ago

8 0

1.

No.

Equation: x-25=40

Variable represents number of donuts

2.

8/3=x

Variable represents ounces of milk

/ means divided by

3.

50-12=x

Variable represents how many hours you work

Sorry for not doing 4

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Complementary and supplementary angles please answer please

kari74 [83]

Answer:

m<MON = 57

Hope this helps :)

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2 years ago

I need help there’s more than one answer .

myrzilka [38]

Answer:

Step-by-step explanation:

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3 years ago

What is the simplest form of the fraction 120/150?

kolbaska11 [484]

Answer:

4/5

Step-by-step explanation:

120/30=4

150/30=5

120/150=4/5

6 0

2 years ago

Read 2 more answers

A group of friends wanted to compare their average running speeds. They recorded the distance and amount of time each person ran

Sauron [17]

Answer with explanation:

Since there is no information provided on what options to determine the proportional relationship, so I will be demonstrating using examples and telling the method to compare the average running speeds.

Consider a group of 3 friends named Charles, Beck and Anna.

As they wanted to compare their average running speeds, they recorded the distance and amount of time each person ran one saturday morning.

To calculate Average Speed, we use:

Average Speed = $\frac{Distance}{Time}$

Example:

Charles Ran:

5 km in 30 minutes

Beck Ran :

2 km in 25 minutes

Anna Ran:

4 km in 20 minutes

Average speed of Charles = $\frac{5(1000)}{30(60)}= 2.78 m/s$

Average speed of Beck = $\frac{2(1000)}{25(60)}= 1.33m/s$

Average speed of Anna = $\frac{4(1000)}{20(60)}= 3.33m/s$

Here, we can see from comparison that Average running speed of Anna is the most.

Speed of Beck is proportional to both Charles and Anna.

Charles' Speed =2 (Beck's Speed)

Anna's Speed =2.5 (Beck's Speed)

6 0

3 years ago

The weights of certain machine components are normally distributed with a mean of 8.5 g and a standard deviation of 0.09 g. Find

stiks02 [169]

To solve this problem,lets say that

X = the weight of the machine components. 
X is normally distributed with mean=8.5 and sd=0.09

We need to find x1 and x2 such that
P(X<x1)=0.03 and P(X>x2)=0.03

Standardizing:

P( Z< (x1 - 8.5)/0.09 ) =0.03
P(Z > (x2 - 8.5)/0.09 ) =0.03.

From the Z standard table, we can see that approximately P = 0.03 is achieved when Z equals to:

z = -1.88 and z= 1.88

Therefore,

P(Z<-1.88)=0.03 and P(Z>1.88)=0.03 

So,

(x1 - 8.5)/0.09 = -1.88 and
(x2 - 8.5)/0.09 =1.88

Solving for x1 and x2:

x1=-1.88(0.09) + 8.5 and
x2=1.88(0.09) + 8.5

Which yields:

x1 = 8.33 g
x2 = 8.67 g

Answer: The bottom 3 is separated by the weight 8.33 g and the top 3 by the weight 8.67 g.

8 0

3 years ago