All categories

4 years ago

Roderick has a dog that has a mass of 9 kilograms what is the mass of the dog

Mathematics

2 answers:

Colt1911 [192]4 years ago

8 0

What unit do you need it in?

Vesna [10]4 years ago

5 0

19.8416lbs
9000grams
9e+6milligrams
0.009 metric ton

You might be interested in

3 more than the product of 3 and y

Fudgin [204]

Answer:

3y+3

Step-by-step explanation:

pls give brainly

4 0

3 years ago

A furniture store bought a couch for $425 and marked the price up by 70%. Angela bought the couch with a 6% tax. How much did An

Tomtit [17]

Answer:Angela paid 679,05 for the couch

Step-by-step explanation:

425*70/100=29.750

29.750/100= 297,5

297,5+425=722,5

722,5*6/100=4.335/100=43,45

722,5-43,45=679,05

8 0

3 years ago

Read 2 more answers

What is the domain of the function y = 2 VX-5?

tresset_1 [31]

Answer:

<em>R</em>

Step-by-step explanation:

All linear functions have a range and domain of R [ALL REAL NUMBERS].

I am joyous to assist you anytime.

3 0

3 years ago

Please help this is the easiest question ever.

Elodia [21]

The answer to this question is 22

8 0

3 years ago

Read 2 more answers

(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

3 years ago