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3 years ago

Jake has 19 pumpkins in his field. kayla has 47 pumpkins in hers. how many pumpkins do they have all together

Mathematics

2 answers:

nadezda [96]3 years ago

8 0

Answer:

19 pumpkins (Jake) + 47 pumpkins (Kayla)= 66 pumpkins

Step-by-step explanation:

just add both numbers and you will get the total. have a nice day!:)

natima [27]3 years ago

5 0

Answer:

Jake and Kayla has 66 pumpkins in their field all together.

Step-by-step explanation:

(Step 1: add 7 + 9, and you should get 16. Add the 1 on top of the 4, and place the 6 below. )

(Step 2: Add 4+1, plus the 1 on top of the 4, and you should get 6.)

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What is equalvent to 7:15 A 21:45 b 14:45 c 3:5 d65:135

Ivanshal [37]

Answer:

its a 21:45

Step-by-step explanation: it a because 7x3 will equal 21 you got 21 in there. 15 x 3 will equal 45 and you got 45 there so a is your answer.

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2 years ago

How Do You Write The Expression For Triple 9 Then Multiply 8

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2 years ago

Find the distance between the pair of points and then round your answer to the nearest tenth.

Travka [436]

Answer:

$d = 9.4 units$ (nearest tenth)

Step-by-step explanation:

Given the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , distance between (-4, 5) and (4, 0) is calculated as follows:

Let $(-4, 5) = (x_1, y_1)$

$(4, 0) = (x_2, y_2)$

$d = \sqrt{(4 - (-4))^2 + (0 - 5)^2}$

$d = \sqrt{(8)^2 + (- 5)^2}$

$d = \sqrt{64 + 25}$

$d = \sqrt{89}$

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3 0

3 years ago

Rosario draws the right triangular prism shown here and calculates the volume. She then draws a second right triangular prism in

Tju [1.3M]

The volume of the original prism is given by:
V1 = (1/2) * (b) * (h) * (H)
Where,
b: base of the triangle
h: height of the triangle
H: prism height
The volume of the prism with new dimensions is:
V2 = (1/2) * (3b) * (3h) * (3H)
Rewriting:
V2 = (3 * 3 * 3) * (1/2) * (b) * (h) * (H)
V2 = (27) * (1/2) * (b) * (h) * (H)
V2 = (27) * V1
Answer:
The relationship between the volumes of the two prisms is:
V2 = (27) * V1

5 0

3 years ago

Read 2 more answers

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago