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Tems11 [23]
1 year ago
8

A scuba diving instructor takes a group of students to the depth of 54.96 feet. Then they ascend 22.38 feet to see some fish. Wh

ere are the fish in relation to the surface?
Mathematics
2 answers:
saul85 [17]1 year ago
8 0

Answer:

The first relation to the surface = -32.58 feet

Step-by-step explanation:

A scuba diving instructor takes a group of students to the depth of 54.96 feet.

It is represented by -54.96 feet

Then they ascend 22.38 feet to see some fish.

It is represented by +22.38 because they are climb up from the deep water.

So, the first relation to the surface = -54.96 + 22.38

= -32.58

The answer is -32.58 feet

stiv31 [10]1 year ago
6 0
32.58 feet deep
I think
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A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th
GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. 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)}.

In this problem, we have that:

p = 0.22, n = 425

So

\mu = E(X) = np = 425*0.22 = 93.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

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

Z = \frac{82 - 93.5}{8.54}

Z = -1.35

Z = -1.35 has a pvalue of 0.0885.

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

8 0
1 year ago
A local university wanted to understand what students prefer to eat during finals. They asked 1,000 students, "Do you prefer chi
kompoz [17]
Given that the person is female, the universe is reduced to 219 + 192 + 119 = 530 people.

The number of women that prefer pizza is 119.

Then the probability is 119/530 *100 = 22.5%
3 0
2 years ago
Read 2 more answers
A manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly.
denpristay [2]

Answer:

We conclude that the population mean light bulb life is at least 500 hours at the significance level of 0.01.

Step-by-step explanation:

We are given that a manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly. The population standard deviation is 50 hours and the light bulb life is normally distributed.

You select a sample of 100 light bulbs and find mean bulb life is 490 hours.

Let \mu = <u><em>population mean light bulb life.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 500 hours      {means that the population mean light bulb life is at least 500 hours}

Alternate Hypothesis, H_A : \mu < 500 hours     {means that the population mean light bulb life is below 500 hours}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                           T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean bulb life = 490 hours

           σ = population standard deviation = 50 hours

           n = sample of light bulbs = 100

So, <u><em>the test statistics</em></u>  =  \frac{490-500}{\frac{50}{\sqrt{100} } }

                                     =  -2

The value of z test statistics is -2.

<u>Now, at 0.01 significance level the z table gives critical value of -2.33 for left-tailed test.</u>

Since our test statistic is higher than the critical value of z as -2 > -2.33, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis.</u>

Therefore, we conclude that the population mean light bulb life is at least 500 hours.

7 0
1 year ago
Suppose that Kevin can choose to get home from work by car or bus. When he chooses to get home by car, he arrives home after 7 p
astraxan [27]

Answer:

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm = 0.838

Step-by-step explanation:

Let the probability that Kevin arrives home after 7 pm be P(L)

Probability that Kevin uses the bus = P(B)

Probability that Kevin uses the car = P(C)

Probability of arriving home after 7 pm if the car was taken = P(L|C) = 4% = 0.04

Probability of arriving home after 7 pm if the bus was taken = P(L|B) = 15% = 0.15

The bus is cheaper, So, he uses the bus 58% of the time.

P(B) = 58% = 0.58

P(C) = P(B') = 1 - P(B) = 1 - 0.58 = 0.42

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm = P(B|L)

The conditional probability P(A|B) is given mathematically as

P(A|B) = P(A n B) ÷ P(B)

Hence, the required probability, P(B|L) is given as

P(B|L) = P(B n L) ÷ P(L)

But we do not have any of P(B n L) and P(L)

Although, we can obtain these probabilities from the already given probabilities

P(L|C) = 0.04

P(L|B) = 0.15

P(B) = 0.58

P(C) = 0.42

P(L|C) = P(L n C) ÷ P(C)

P(L n C) = P(L|C) × P(C) = 0.04 × 0.42 = 0.0168

P(L|B) = P(L n B) ÷ P(B)

P(L n B) = P(L|B) × P(B) = 0.15 × 0.58 = 0.087

P(L) = P(L n C) + P(L n B) = 0.0168 + 0.087 = 0.1038 (Since the bus and the car are the two only options)

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm

= P(B|L) = P(B n L) ÷ P(L)

P(B n L) = P(L n B) = 0.087

P(L) = 0.1038

P(B|L) = (0.087/0.1038) = 0.838150289 = 0.838

Hope this Helps!!!

8 0
1 year ago
Jacob finished his test 10 minutes before Mark. Which equation represents the time, j, Jacob spent on the test compared to the t
Marina CMI [18]

Answer:

The time Jacob spent on the test represented by J = (m + 10) minutes

where m is the number of minutes spent by Mark on the test

Step-by-step explanation:

Here, we want to give an equation which represents the time taken for Jacob to finish his test compared to the time m which Mark spent on the test.

Since Jacob finished 10 minutes before Mark, it means Mark actually took longer on the test.

Now, we are told that Mark spent m minutes on the test, the time spent by Jacob on the test is thus (m + 10) minutes

Let the amount of time spent by Jacob be J minutes. This in relation to the amount of time spent by Mark will be ;

J = (m + 10) minutes

6 0
1 year ago
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