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2 years ago

Complete the comparison: 26 <

Mathematics

1 answer:

EastWind [94]2 years ago

6 0

26< any number larger than 26 (whole numbers and fractions/decimals)

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Can someone help me out pls

Fiesta28 [93]

the answer is d 350000..

6 0

2 years ago

What is 243.875 rounded to the nearest tenth,hundreth,ten,and hundred?

ira [324]

Tenth: 243.9
hundreth: 243.88
ten: 240
hundred:200

hope this helps

5 0

2 years ago

Transform the following polar equation into an equation in rectangular coordinates: r=2 cos theta A. x + y = 2 B. y = 2x C. x =

Mila [183]

$r=2\cos\theta$
$r^2=2r\cos\theta$
$\implies x^2+y^2=2x$
$x^2-2x+y^2=0$
$x^2-2x+1+y^2=1$
$(x-1)^2+y^2=1$

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

2 years ago

-6d - (c - d)^2 ; if c = 7 and d = 4

irina1246 [14]

Answer:

-6d - (c - d)^2 = -33

Step-by-step explanation:

-6(4) - (3)^2 =

-24 - 9 =

-33

7 0

3 years ago