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3 years ago

5

At a local high school, 30% of the students are on a school sports team, 25% take at least one Advanced Placement (AP) class, an

d
10% are on a sports team and take at least one AP class. If a student was randomly selected from this school, what is the
probability that the student is on a sports team or takes at least one AP class?

Plz hurryyyy

1 answer:

gogolik [260]3 years ago

8 0

Answer:

not sure but 30 over 100 for the sports and 10 over 100 for at least one ap class

Step-by-step explanation:

can u please comment I need to talk to you

hurry

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You and your friend live 12 miles apart. you leave home at the same time and travel ttoward eacth other.you walk at the rate of

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Step 1: Assign variable for the unknown that you need to find.

$d_w \; \rightarrow \; Distance \; you \; traveled \; by \; walk \; in \; miles \\ d_c \; \rightarrow \; Distance \; traveled \; by \; your \; friend \; using \; bicycle \; in \; miles \\ t \; \rightarrow \; Time \; Taken \; in \; hours$

Step 2: List the given details

Distance between you and your friend is 12 miles. It means the total distance traveled by you by walk and your friend using bicycle when both travel towards each other to meet in 't' hours

Equation 1:

$d_w+d_c=12$

Your Walking rate and your friends Cycling rate is given below:

$r_w=4mph\\r_c=11mph$

Step 3: Set up an equation using the relationship between distance, time and rate.

$Distance=Rate \times Time$

$d_w=r_w \times t=4t \\\\ d_c = r_c \times t = 11t$

Substituting the above in equation 1

$4t+11t=12 \\ Combining \; Like \; Terms \\\\ 15t=12 \\ Dividing \; 15 \; on \; both \; sides \\\\ 15t \div 15=12 \div 15 \\\\ t=0.8 hours$

Distance traveled by you by walking from home and distance travelled by your friend in cycle in 0.8 hours is given below:

$d_w=4t=4(0.8)=3.2miles \\\\ d_c=11t=11(0.8)=8.8miles$

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3 years ago

Six percent (6%) of all shoppers make an impulse purchase at the check-out counter in a grocery store. If a store has 650 custom

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Answer:

(e) 39

Step-by-step explanation:

The expected value (or the average number) of impulse purchases per day is given by the probability of an impulse purchase being made (6%) multiplied by the daily number of customers (650):

$E(X) =P(X)* n\\E(X) = 0.06*650\\E(X) = 39$

The average number of impulse purchases is 39 per day.

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3 years ago

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3 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

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3 years ago