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asambeis [7]
3 years ago
10

Can you help me with 10-13 please? I don’t really understand it

Mathematics
1 answer:
Verdich [7]3 years ago
5 0
The answer 0 10_13 --0 zero
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A copy machine makes 28 copies per minute. How many copies does it make in 3 minutes and 45 seconds?
lina2011 [118]
<u>Convert minutes and seconds to minutes:</u>
\text{3 mins 45 seconds} \longrightarrow 3 \dfrac{3}{4} \text{mins}


<u>FInd 3 3/4 mins:</u>
\text{1 min} \longrightarrow 28 \text{ copies}

3  \dfrac{3}{4} \text{ min} \longrightarrow 28 \times 3\dfrac{3}{4} \text{ copies}

3  \dfrac{3}{4} \text{ min} \longrightarrow 28 \times \dfrac{15}{4} \text{ copies}

3  \dfrac{3}{4} \text{ min} \longrightarrow 105 \text{ copies}

Answer: It can make 105 copies.

4 0
4 years ago
A physical fitness association is including the mile run in its secondary-school fitness test. The time for this event for boys
Luba_88 [7]

Answer:

z=-1.28

And if we solve for a we got

a=470 -1.28*60=393.2

So the value of height that separates the bottom 10% (Fastest 10%) of data from the top 90% is 393.2.

so if the time is 393.2 or lower a boy would earn a certificate of recognition from the fitness association

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the time for this event of a population, and for this case we know the distribution for X is given by:

X \sim N(470,60)  

Where \mu=470 and \sigma=60

NOTE: For this case the fastest 10% is on the lower tail of the distribution

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.9   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-1.28

And if we solve for a we got

a=470 -1.28*60=393.2

So the value of height that separates the bottom 10% (Fastest 10%) of data from the top 90% is 393.2.

so if the time is 393.2 or lower a boy would  earn a certificate of recognition from the fitness association

4 0
3 years ago
What is the product of 4.5 × 10-5 and 2.4 × 10-2?
nikitadnepr [17]
4.5 • 10-5 = 40

2.4 • 10-2 = 22
4 0
3 years ago
Read 2 more answers
What is the length of the hypotenuse of the triangle when x=15?
Luden [163]

Answer:

114.24°

Step-by-step explanation:

7(15)+6= 105

3(15)= 45, now we plug it in the formula, a^{2}+b^{2}=c^{2}

105^{2}+45^{2} = 13050

so now we square root it

\sqrt{13050}= 114.24

6 0
3 years ago
Read 2 more answers
Marlene went shopping and bought a bunch of candy for her swim team. She bought packs of Skittles for $1.50 each and packs of M&
lord [1]

Answer:

18 Skittles

6 M&Ms

Step-by-step explanation:

Set up an equation:

Variable x = number of skittles

Variable y = number of M&Ms

1.50x + 2y = 39

x + y = 24

In the second equation, isolate a variable:

x = 24 - y

Substitute the value of x for 24 - y in the first equation:

1.50(24 - y) + 2y = 39

Use distributive property

36 - 1.5y + 2y = 39

Combine like terms

36 + 0.5y = 39

Isolate variable y:

0.5y = 3

y = 6

Substitute the value of y for 6 in the second equation:

x + 6 = 24

Isolate variable x:

x = 18

Plug these values into any equation of your choice to see if these values are correct (I'll do both equations just to prove it):

1.50(18) + 2(6) = 39

27 + 12 = 39

39 = 39

Correct

x + y = 24

18 + 6 = 24

24 = 24

Correct

8 0
3 years ago
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