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Nutka1998 [239]
3 years ago
10

Sean answered 18 of 20 quiz questions correctly. What percent of the quiz questions did Sean answer correctly?

Mathematics
2 answers:
ozzi3 years ago
8 0
The answer is 90%. You get this answer by multiplying both the numerator and denominator by 5 so 18x5=90 and 20x5=100 giving you a fraction of 90/100 and a percentage of 90%.
ipn [44]3 years ago
4 0

Answer:

Sean answered 90% of the questions in the quiz correctly.

Step-by-step explanation:

Sean answered 18 of 20 quiz questions correctly.

So, to know the percentage we can do the following calculation.

\frac{18}{20}\times100= 90%

Therefore, Sean answered 90% of the questions in the quiz correctly.

You might be interested in
A line with a slope of -2 is written in the form y=mx+b.
yanalaym [24]

Answer:

b = 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + b ( m is the slope and b the y- intercept )

Here m = - 2 , then

y = - 2x + b ← is the partial equation

To find b substitute (5, - 7 ) into the partial equation

- 7 = - 10 + b ⇒ b = - 7 + 10 = 3

5 0
2 years ago
29
kramer

Answer: C (1.25, 10.25)

Step-by-step explanation:

4 0
3 years ago
How do you do this question?
Alex Ar [27]

Step-by-step explanation:

(a) dP/dt = kP (1 − P/L)

L is the carrying capacity (20 billion = 20,000 million).

Since P₀ is small compared to L, we can approximate the initial rate as:

(dP/dt)₀ ≈ kP₀

Using the maximum birth rate and death rate, the initial growth rate is 40 mil/year − 20 mil/year = 20 mil/year.

20 = k (6,100)

k = 1/305

dP/dt = 1/305 P (1 − (P/20,000))

(b) P(t) = 20,000 / (1 + Ce^(-t/305))

6,100 = 20,000 / (1 + C)

C = 2.279

P(t) = 20,000 / (1 + 2.279e^(-t/305))

P(10) = 20,000 / (1 + 2.279e^(-10/305))

P(10) = 6240 million

P(10) = 6.24 billion

This is less than the actual population of 6.9 billion.

(c) P(100) = 20,000 / (1 + 2.279e^(-100/305))

P(100) = 7570 million = 7.57 billion

P(600) = 20,000 / (1 + 2.279e^(-600/305))

P(600) = 15170 million = 15.17 billion

7 0
2 years ago
Math:<br> Please help me I don’t understand and I need to do corrections.
GREYUIT [131]
So, you had done everything right so far (other than squaring the 2), but that was only half of the question.

to find the least common multiple, you need to first figure out what the prime factors have in common.
{2}^{2}  \times 3 \times 5 \\ and \\  {2}^{2}  \times  {3}^{2}  \times 5 \times 7
each have two twos. both have one 5, so we know our answer will look something like
{2}^{2}  \times 5 \times other \: stuff
now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.

one has one 3 and one has two, so we need two threes. now our equation is
{2}^{2}  \times {3}^{2}  \times 5 \times stuff

what's the only number we have to deal with? 7...

how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes
4 0
3 years ago
Hey you can help help pls pls help help me
AleksandrR [38]

Answer:

A.)

Step-by-step explanation:

HOPE IT HELPED:)!!

4 0
2 years ago
Read 2 more answers
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