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3 years ago

Which situation is most likely to have a constant rate of change

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

8 0

The answer is:

B. The cost for a group of friends to see a movie compared to the number of friends.

We can tell that this is most likely to have a constant rate of change because of what we already know about movie prices.

Movie prices are fixed, therefore, no matter how many friends you bring, the total price will always go up by the same amount.

For example:

Let's say a movie ticket costs $5 and you brought 3 friends with you. Since there are 4 people in total, you'd have to pay $20 for the tickets (5x4=20). Just as you're about to pay, Sally shows up. Now you have to pay for 5 people in total. How much do you need to pay now?

You will now need to pay $25 (5x5=25) for all of the tickets. Even though you now have 5 people in the group, you still only have to pay $5 for each ticket.

This is an example of a constant rate of change.

Karo-lina-s [1.5K]3 years ago

7 0

B i agree with the other person i believe

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KiRa [710]

∠1 and ∠3 complementary,
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∠1 and ∠2 complementary.

Answer : C.

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3 years ago

In 2014 Rose invested $16,000 in a savings account for her newborn son. The account pays 3.6% interest each year. Determine the

Juliette [100K]

Answer:

Total value of the account in 2032 will be $26,368

7 0

2 years ago

Ill give brainlist

Mariana [72]

The missing step 2 should be;

The quantity of the sum of thirty six plus two divided by 0.5 − 14.8 ÷ 8

The missing step 6 would be that; 76 - 1.85

<h3>How to Solve Mathematical Order of Operations?</h3>

We want to know the steps to solving the expression the quantity of the sum of six squared plus two divided by the absolute value of negative 0.5 − 14.8 ÷ 8

This can be expressed as;

6² + 2 ÷ |-0.5| - 14.8 ÷ 8

Thus, the next step which is step 2 should be;

The quantity of the sum of thirty six plus two divided by 0.5 − 14.8 ÷ 8

The missing step 6 would be that;

76 - 1.85

Read more about Order of Operations at; brainly.com/question/27529825

#SPJ1

7 0

2 years ago

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr

sergij07 [2.7K]

$f(t)=\begin{cases}0&\text{for }t$

Recall that

$u(t)=\begin{cases}0&\text{for }t$

Take it one piece at a time. For $t\ge0$ , we can scale $u(t)$ by -5:

$-5u(t)=\begin{cases}0&\text{for }t$

If we shift the argument by 1 and scale by -5, we have

$-5u(t-1)=\begin{cases}0&\text{for }t$

so if we subtract this from $-5u(t)$ , we'll end up with

$-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t$

For the next piece, we can add another scaled and shifted step like

$-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

so that

$-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

For the last piece, we add one more term:

$6u(t-7)=\begin{cases}0&\text{for }t$

and so putting everything together, we get $f(t)$ :

$f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)$
$f(t)\equiv-5u(t)-u(t-1)+12u(t-7)$

5 0

3 years ago

The Town of Hertfordshire clerk knows that 23% of dogs in the town have completed emotional support training. Hertfordshire plan

Nataly_w [17]

Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .