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tino4ka555 [31]
3 years ago
13

Three of these turtles weight 1,050 pounds. How much does one Pinta Island Tortoise weight?

Mathematics
2 answers:
aleksklad [387]3 years ago
8 0

Answer:

350 pounds

Step-by-step explanation:

Assuming three of the turtles are Pinta Island Tortoise, then:

3 turtles -- 1050

1 turtle -- 1050/3 = 350

Thenks and mark me brainliest :))

sladkih [1.3K]3 years ago
6 0

Answer:

the answer is 350

Step-by-step explanation:

1050 divided by 3 equals 350!!! hope this helps

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I wish to have $10,000 at the end of 8 years in a bank offering a simple interest rate of 7.5% per year. How much should I depos
cluponka [151]

Answer:

$6250

Step-by-step explanation:

6250x.075x8=3750

6250+3750=10,000

7 0
3 years ago
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
3 years ago
For a data set of the pulse rates for a sample of adult? females, the lowest pulse rate is 39 beats per? minute, the mean of the
Zarrin [17]

Answer:

a) The difference is of 35 beats per minute.

b) So 3.07 standard deviations below the mean.

c) Z = -3.07.

Step-by-step explanation:

Z-score:

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean \mu = 74, standard deviation \sigma = 11.4

a. What is the difference between the pulse rate of 39 beats per minute and the mean pulse rate of the? females?

39 - 74 = -35

The difference is of 35 beats per minute.

b. How many standard deviations is that? [the difference found in part? (a)?

|Z| when X = 39. So

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

Z = \frac{39 - 74}{11.4}

Z = -3.07

|Z| = 3.07

So 3.07 standard deviations below the mean.

c. Convert the pulse rate of 39 beats per minutes to a z score.

From item b. above, Z = -3.07.

5 0
2 years ago
Eddie was watching a movie with a total length of 3 3/5 hours. He watched 1 1/4 hours of the movie but paused it to do his chore
Rudik [331]
1hr 6min left because all you have to do is keep subtracting       
6 0
3 years ago
Write the vector u as a sum of two orthogonal vectors, one of which is the vector projection of u onto v, projvu. u = <-8, -8
aalyn [17]

Answer:

c.) <-1, 2> + <-2, -1>

Step-by-step explanation:

hope i helped

3 0
3 years ago
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