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3 years ago

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A deep sea diver started at sea level (0 feet) and dove at a rate of 5.6 feet per minute. If the diver dove for ½ hour, what rat

ional number represents his position relative to sea level? Show your work. A. 168 feet B. 2.8 feet C. -2.8 feet D. -168 feet

1 answer:

Leona [35]3 years ago

5 0

Given:

A deep sea diver started at sea level (0 feet).

Rate of diving = 5.6 feet per minute.

Time = $\dfrac{1}{2}$ hour

To find:

The rational number that represents his position relative to sea level.

Solution:

We know that,

1 hour = 60 minutes

$\dfrac{1}{2}$ hour = 30 minutes

A deep sea diver, dove at a rate of 5.6 feet per minute.

Distance covered in 1 minute = 5.6 feet

Distance covered in 30 minute = 30 × 5.6 feet

= 168 feet

The diver dove 168 feet in $\dfrac{1}{2}$ hour. Since it is below the sea level, therefore it should be negative.

So, the rational number that represents his position relative to sea level is $-168$ feet.

Therefore, the correct option is D.

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Step-by-step explanation:

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3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

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3 years ago

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Ne4ueva [31]

Answer:

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Step-by-step explanation:

Pink Rectangle:

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3 years ago

The probability that a randomly selected 2 2-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0

Mnenie [13.5K]

Answer:

a. $Probability = 0.97735$

b. $Probability = 0.92294$

c. $P(At\ Least\ One) = 1$

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

$P(Live) = 0.98861$

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)\ and\ P(Live)$

$Probability = 0.98861 * 0.98861$

$Probability = 0.9773497321$

$Probability = 0.97735$ <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)^n$

Where $n = 7$

$Probability = 0.98861^7$

$Probability = 0.92294324145$

$Probability = 0.92294$ <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

$P(At\ Least\ One) = 1 - P(None).$

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

$P(Not\ Live) = 0.01139$

The probability that none of the 7 lives up to 3 is:

$P(None) = P(Not\ Live)^7$

$P(None) = 0.01139^7$

Substitute this value for P(None) in

$P(At\ Least\ One) = 1 - P(None).$

$P(At\ Least\ One) = 1 - 0.01139^7$

$P(At\ Least\ One) = 0.99999999999997513055642436060443621$

$P(At\ Least\ One) = 1$ ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

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3 years ago

What are the zeros of the function y = 2x2 + 5x + 2?

Dvinal [7]

Answer: not sure

but I need points

Step-by-step explanation:

JUST KIDDING It’s A I did this in school before and got it correct :)

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3 years ago