Please help me :( with this

Three tickets to a movie costs $37.50. If all tickets were the same price, what would be the cost of 2 tickets?

Anon25 [30]

Answer:

$25 I believe

Step-by-step explanation:

You would take $37.50 divided by 3 and get $12.50 which is the price of 1 ticket

Then you would multiply $12.50 x 2 to get $25

4 0

3 years ago

Read 2 more answers

Ramon earns $1,880 each month and pays $53.30 on electricity. To the nearest tenth of a percent, what percent of Ramon's earning

AfilCa [17]

<h2>Answer:</h2>

Each month, 2.8% Ramon's earnings are spent on electricity.

<h2>Step-by-step explanation:</h2>

You know, that 100 percent are his earnings - $1,880

Now, you need to find out, how much is 1%. You do that by dividing $1,880 by 100.

Now you have to divide the amount he pays on electricity - $53.3 by one percent of his earnings - $18.8

So, now you know, that he pays exactly 2.83511% of his earnings on electricity. But from assignment, you know, that it has to be rounded to the nearest tenth of a percent. The number is 2.8351. So we will round it to 2.8% ,because 3 is rounded down. (https://www.mathsisfun.com/rounding-numbers.html)

100% = $1,880

$1,880/100 = $18.8

$53.3/$18.8 = 2.83511

2.83511% ≈ 2.8%

3 0

2 years ago

One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a

igor_vitrenko [27]

Answer: $P=0.975^t$

Step-by-step explanation:

We know that the general form of the exponential decay formula is

$y=A(1-r)^t$ , where y is final amount remaining after t time, A is the original amount and r is the rate of decay

Now, the ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident. $P=\frac{A(1-0.02)^t}{A}=\frac{(0.975)^t}{1}$

Hence, the ratio of remaining since the nuclear accident is $P=0.975^t$

3 0

2 years ago

Read 2 more answers

Which of the following arguments correctly justifies the second step of the flow proof of the triangle angle sum theorem below

pochemuha

We need to see the second step of the proof, so we know what needs to be justified. It would also be nice to see the list of answer choices. I'll never understand how someone who can't answer the question and has to ask someone else can then go ahead and decide that the person helping him only needs to see parts of the question.

8 0

2 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

2 years ago