Test worth 50 points; test score of 78%?

Mathematics

1 answer:

nataly862011 [7]3 years ago

7 0

The person would have scored 39 out of 50. Hope this helps

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Probability is used when describing a function of a parameter given an outcome. For example, if a coin is flipped 10 times and i

Paladinen [302]

Answer: 1/1024

Step-by-step explanation:

When a fair coin is flipped the coin would show either head or tail meaning only 2 outcomes are possible. So, getting an head means 50% chance or 1/2 chance.

Now, it means chance for 10 times head every time= (1/2)^10

Chance= 1/1024

Chance= 0.00097

8 0

3 years ago

If KM = NO and NO = PQ, then KM = PQ.

uysha [10]

The answer is D. transitive property (algebra lesson)

4 0

4 years ago

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What is -5 3/4 - 3 1/2 =

Luba_88 [7]

Answer:

-9 1/4

Step-by-step explanation:

-5 3/4 - 3 1/2

-5 3/4 - 3 2/4

-9 1/4

7 0

2 years ago

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Which strategy best explains how to solve this problem?

Rufina [12.5K]

This is a geometric sequence because each term is twice the value of the previous term. So this is what would be called the common ratio, which in this case is 2. Any geometric sequence can be expressed as:

a(n)=ar^(n-1), a(n)=nth value, a=initial value, r=common ratio, n=term number

In this case we have r=2 and a=1 so

a(n)=2^(n-1) so on the sixth week he will run:

a(6)=2^5=32

He will run 32 blocks by the end of the sixth week.

Now if you wanted to know the total amount he runs in the six weeks, you need the sum of the terms and the sum of a geometric sequence is:

s(n)=a(1-r^n)/(1-r) where the variables have the same values so

s(n)=(1-2^n)/(1-2)

s(n)=2^n-1 so

s(6)=2^6-1

s(6)=64-1

s(6)=63 blocks

So he would run a total of 63 blocks in the six weeks.

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3 years ago

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Earth revolves around the Sun in 365 days, but the planet Mercury does so in only 88 days. Compared to Earth, how many more comp

Elena L [17]

Answer:

In a 1000 days Mercury will make 8.624 more revolutions than Earth!

Step-by-step explanation:

Earth takes 365 days to make a complete revolution.

we can say that the rotational speed of Earth is $\dfrac{1}{365}\frac{\text{rev}}{\text{day}}$