I dont get how to do this one

A large dump truck uses 1,318 gallons of fuel in one day of work. How many gallons of fuel are needed if the truck works for 5 d

Andreyy89

Answer:

6,590 Gallons

Step-by-step explanation:

1,318 x 5 = <em>x</em>

<em>x</em> = 6,590

5 0

3 years ago

Read 2 more answers

Find the sum of 9, eight times a number y and five times

MatroZZZ [7]

Step-by-step explanation:

Fill in: turn into, access, instantly, endless,

dissolve, assembly.

1. Spray-on clothes contain minute fibres which

dry........................... .

2 The Airbike is ready to ride as it doesn't require

any........................... .

3 Dr. Torres has developed a fabric that can .................... any

garment.

4. The possibilities of using the new spray are....................... .

5. A touchscreen gives you instant......................... to the Internet.

6. Some supermarkets are using plastic bags which.....................in water, leaving no trace.

7 0

3 years ago

Amber has 5/2 pounds of salt dough for a project. she writes this equation to show 5/2. 5/2= 5×1/2

Alona [7]

If you would like to know whether or not what she wrote is accurate, it is.

Hope that helps!

3 0

3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

What is the answer to 2/3j-1/6=1/2?

Elodia [21]

It has already given you the answer.

3 0

3 years ago