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3 years ago

How many millilitres are there in 51%

Mathematics

1 answer:

o-na [289]3 years ago

4 0

Answer: 51 oz is equal to 1,508.27 ml, or there are 1,508.27 milliliters in 51 ounces.

HOPE THIS HELPS

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8 8/9 - 5 1/11???????????????????????

saw5 [17]

Hiiii

Answer:

The answer is 4 25/99

$4 \frac{25}{99}$

5 0

2 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

This is worth 18 points to whoever gets it right and also if you do get it right first you'll get voted as brainlyess answrr

Bumek [7]

The correct answer is <span>1 and 1/2

Hope I helped</span>

4 0

3 years ago

Read 2 more answers

When solving the equation such as 2/3x = 4/5 , why would you multiply both sides of the equation by the reciprocal of 2/3 ?

barxatty [35]

Answer: Well you’re solving for “x,” so you’ll need to isolate this variable. This means we need to “get rid of” the fraction 2/3 that is attached to it. The faction is attached by the operation multiplication, so to remove it we must divide it on both sides. And division with fractions requires the reciprocal. (Try watching an explanation if you’re confused on why division requires reciprocals).

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3 years ago

Trigonometry and Geometric Modeling; the ratios, which are called trigonometric ratios, include __________, ____________, and __

Margarita [4]

Answer:

Sine, cosine and tangent

Step-by-step explanation:

Given a right triangle you can find the sine (or cosine, or tangent) of either of the non- 90° angles.

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3 years ago