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4 years ago

5

You are watching your favorite soccer player on TV. Statistics show that he has a 37% chance of scoring a goal each time he shoo

ts. He shoots twice. What is the probability that he scores a goal both times?

1 answer:

e-lub [12.9K]4 years ago

7 0

Likely and unlikely is the answer

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When solving for a system of equations, if you get an answer of 6=0, what does this solution mean regarding the solution to the

lara [203]

It means the system is inconsistent, which is another way to way, there's no solution, namely, both graphs do not intersect ever, and therefore there's no common point with such x,y pair for both.

8 0

3 years ago

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Joanna ate 3/7 of a pizza . Michael ate 2/7 of the pizza

SpyIntel [72]

Answer:

5/7

Step-by-step explanation:

3+2 is 5 over the same denominator because that's how many slices of pizza there are

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2 years ago

A population of bacteria is growing exponentially. At 7:00 a.m. the mass of the population is 12 mg. Five hours later it is 14 m

sergejj [24]

Answer with Step-by-step explanation:

The exponential growth function is given by

$N(t)=N_oe^{mt}$

where

$N(t)$ is the population of the bacteria at any time 't'

$N_o$ is the population of the bacteria at any time 't = 0'

'm' is a constant and 't' is time after 7.00 a.m in hours

Assuming we start our measurement at 7.00 a.m as reference time t = 0

Thus we get $N(0)=N_oe^{m\times 0}\\\\12=N_o$

Now since it is given after 5 hours the population becomes 14 mg thus from the above relation we get

$12\times e^{m\times 5}=14\\\\e^{5m}=\frac{14}{12}\\\\m=\frac{1}{5}\cdot ln(\frac{14}{12})\\\\m=0.031$

Thus the population of bacteria at any time 't' is given by

$N(t)=12e^{0.031t}$

Part a)

Population of bacteria after another 5 hours equals the population after 10 hours from start

$N(10)=12e^{0.031\times 10}=16.361mg$

Part b)

Population of bacteria at 7:00 p.m is mass after 12 hours

$N(1)=12e^{0.031\times 12}=17.41mg$

Part c)

Population of bacteria at 8:00 p.m is mass after 1 hour

$N(1)=12e^{0.031\times 1}=12.3378mg$

Part d)

Differentiating the relation of population with respect to time we get

$N'(t)=\frac{d(12\cdot e^{0.031t})}{dt}\\\\N'(t)=12\times 0.031=0.372e^{0.031t}$

Thus we can see that the percentage increase varies with time initially the percentage increase is 37.2% but this percentage increase increases with increase in time

Part 4)

Since there are 24 hours in 1 day thus the percentage increase in the population is

$\frac{N(24)-N_o}{N_o}\times 100\\\\=\frac{25.25-12}{12}\times 100=110.42$

Thus there is an increase of 110.42% in the population each day.

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4 years ago

SOMEONE PLS HELP QUICK

Ket [755]

2. NOT A GOOD FIT

2. RANDOM

4 0

3 years ago

Read 2 more answers

Use the unit circle to find the exact value of the trig function cos(210)

docker41 [41]

Answer:

Step-by-step explanation:

cos(210)=cos(180+30)=-cos 30

$=-\frac{\sqrt{3} }{2}$

7 0

3 years ago