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2 years ago

North lake High school has two lunch periods.... select the true statement

Mathematics

2 answers:

andrew11 [14]2 years ago

5 0

Answer:

Option D.

Step-by-step explanation:

Northlake high school has two lunch periods. Students can eat their lunch in the cafeteria or on an outside patio.

Percentage of students who have first lunch eat outside= $\frac{\text{Eat outside}}{\text{Total of first lunch}}$ = $\frac{(.17)(100)}{.55}=31$ %

Similarly percentage of second lunch students eat outside= $\frac{\text{Eat outside}}{\text{Total of second lunch}}$ = $\frac{(0.21)(100)}{(0.45)}=47$ %

Therefore, option D is the correct option.

Snowcat [4.5K]2 years ago

3 0

Answer: D

Step-by-step explanation:

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2 years ago

Determine if the statement is true or false:

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1 year ago

Prompt:

Vadim26 [7]

Answer:

Her answer is wrong because she the object cannot hit the ground at negative seconds. She could’ve have used other methods because she used the quadratic formula. The advantages is that it works for every situation. The disadvantages is that it takes longer. She should’ve used a different method.

Step-by-step explanation:

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2 years ago

Read 2 more answers

1. Write an exponential function to represent the spread of Ben's social media post. 2. Write an exponential function to represe

Alex777 [14]

Answer:

1. y = 2(3)ˣ; 2. y = 10(4)^x; 4. Amber = 192 and 3 145 728;

Ben has 54 and 11 098; Carter = 80 and 10 249 shares;

5. Shifted up 45 shares; 6. Amber's;

7. Fewer friends. more shares.

Step-by-step explanation:

The general equation for an exponential function is

y = a(b)ˣ

1. Ben's function

The initial value a = 2.

6 = 2(b)¹ = 2b

Ben's function is

y = 2(3)^x

2. Carter's function

Carter shared his post with 10 friends on Day 0, so a= 10.

Each friend shared with 2 people each day, so b = 2.

Therefore, Carter's function is

y=10(2)ˣ

3. Graphs

$\begin{array}{cccc}& \textbf{Amber} &\textbf{Ben} & \textbf{Carter}\\\mathbf{x} & \mathbf{3(4)^{x}} &\mathbf{2(3)^{x}} & \mathbf{10(2)^{x}}\\0 & 3 & 2 & 10\\1 & 12 & 6 & 20\\2 & 48 & 18 & 40\\3 & 192 & 54 & 80\\\end{array}$

The figure below shows the graphs for each function.

4. Predictions

(a) Amber

(i) Day 3

y = 3(4)³ = 3 × 64 = 192

(ii) Day 10

y = 3(4)¹⁰ = 3 × 1 048 576 = 3 145 728

(b) Ben

(i) Day 3

y = 2(3)³ = 2 × 27 = 54

(ii) Day 10

y = 2(3)¹⁰ = 2 × 29 049 = 118 098

(c) Carter

(i) Day 3

y = 10(2)³ = 10 × 8 = 80

(ii) Day 10

y =10(2)¹⁰ = 10 × 1024 = 10 240

5. Sharing with assisted living facility

y = 3(4)ˣ + 45

The new graph will be shifted up by 45 units.

6. Fastest post

Amber's post travels the fastest.

After only two days, she already has more posts than anyone else.

By Day 10, she has more than 25 times the shares of her closest competitor.

7. Preference

It's not the number of friends you have, but the number of shares they make, that is the important factor.

Carter started with 10 friends, but they made only two shares.

Amber started with only three friends, but they made four shares. By Day 3, she had 192 shares compared to 82 for Carter and only 54 for Ben

If I had to choose, I would prefer a post with fewer friends initially but more shares, like Amber.

5 0

2 years ago

g In a survey of 500 residents, 300 were opposed to the use of the photo-cop for issuing traffic tickets.The standard error of t

Evgesh-ka [11]

Answer:

Z-score 1.96

Margin of error d= 0.04323

Step-by-step explanation:

Hello!

The study variable in this case is

X: Number of people that oppose the use of photo-cop for issuing traffic tickets in a sample of 500.

n= 500

The parameter of interest is the proportion of people that opposes it.

The estimated proportion is p'= 300/500= 0.6

To estimate the population proportion per Confidence interval you have to approximate the distribution of the sample proportion p' to normal:

p'≈N(p;p(1-p)1/n)

The mean of this distribution is p and the variance is p*(1-p)*1/n

To construct the Confidence interval, since the value of the population proportion is unknown, an estimated variance is used:

p'(1-p')*1/n ⇒ then the estimated standard error is √(p'(1-p')*1/n)

The formula for the confidence interval is:

p' ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'(1-p')}{n} }$

Where " $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'(1-p')}{n} }$ " represents the margin of error of the interval.

Now for a confidence level of 0.95 the value of Z is $Z_{0.975}= 1.965$

The estimated standard error is already calculated: $\sqrt{\frac{p'(1-p')}{n} } = 0.022$

The margin of error (d) is then:

d= $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'(1-p')}{n} }$ = 1.965 * 0.022

d= 0.04323

I hope it helps!

7 0

1 year ago