Ask question

Ask question

All categories

faltersainse [42]

2 years ago

7

The green turtle lays eggs that are approximately spherical with an average diameter of 4.5 centimeters. each turtle lays an ave

rage of 113 eggs at one time. find the total volume of these eggs, to the nearest cubic centimeter.

1 answer:

Art [367]2 years ago

4 0

The average volume if these eggs will be found as follows:
Volume of one egg is:
V=4/3πr²
V=4/3×π×(4.5/2)^3
V=71.57 cm³
Given that the turtle lays on average 113 eggs, thus the total volume of eggs will be:
(113×71.57)
=8187.41~8187 cm³

You might be interested in

Marks age, X is six times his age 2 years ago. which equation represents the statement

FinnZ [79.3K]

x=6x-2 This would be your answer, hope this helps if you need any further explanation feel free to let me know and I would be happy to help you out.

6 0

2 years ago

Read 2 more answers

Jennifer has a square bedroom. The area of her room is 144 ft2. What is the length of one side of Jennifer’s room?

Jobisdone [24]

Answer:

Option B. $12\ ft$

Step-by-step explanation:

we know that

The area of a square is equal to

$A=x^{2}$

where

x is the length side of the square

In this problem we have

$A=144\ ft^{2}$

substitute and solve for x

$144=x^{2}$

suqre root both sides

$x=12\ ft$

7 0

2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

8 0

2 years ago

The graph shows the level of ibuprofen, y units, in a patient’s bloodstream x hours after the ibuprofen was taken. Select the co

ElenaW [278]

The graph is missing, so i have attached it.

Answer:

From 0 hours to 2 hours.

Step-by-step explanation:

From the attached graph, we can see that on the y-axis, denotes the level of ibuprofen in the patient’s bloodstream while the x-axis denotes the amount of hours after the ibuprofen was taken.

Now, looking at the graph, we can see that the curve goes up from a time of 0 hours which corresponds to 0 mg of ibuprofen to a time of 2 hours which corresponds to 400 mg of ibuprofen.

After that point, the curve begins to to go down. Since we are concerned with increase, we will make do with the first statement.

Thus, we can say the period at which the level of ibuprofen increased was from 0 hours to 2 hours.

6 0

2 years ago