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tamaranim1 [39]

2 years ago

14

Andrew ate 2/3 of the desert. cassandra ate 5/16 of the desert. what portion of the desert did andrew and cassandra eat altogeth

er?

2 answers:

olga2289 [7]2 years ago

8 0

Answer:

The portions added together make 47/48 of the dessert.

Step-by-step explanation:

To find the total amount that Andrew and Cassandra ate of the dessert, we have to add the two fractions together:

$\frac{2}{3} + \frac{5}{16}$

$\frac{32}{48} + \frac{15}{48}$

$\frac{47}{48}$

Hence, Andrew and Cassandra almost ate up all of the dessert, clocking in at 47/48 of the dessert.

Hope this helped!

Diano4ka-milaya [45]2 years ago

4 0

Answer:

Concept: Fractions

A ate 2/3 and C ate 5/16
Let us find a common denominator
Let 48 be our common D
Hence 32/48+ 15/48 = 47/48

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(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

Please help with this question, thank you :)

vovikov84 [41]

Answer:

1/2

((-2)^2-(4*2)^1/3)/abs(-2*2)

-2^2 = 4

4*2 = 8

-2*2 = -4 and abs of that is 4

4-(8)^1/3/4

8^1/3 = 2

4-2 = 2

2/4 = 1/2

Step-by-step explanation:

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ruslelena [56]

Answer:

Step-by-step explanation:

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avanturin [10]

Answer:

Changes made to your input should not affect the solution:

(1): "^-1" was replaced by "^(-1)".

STEP

1

:

Final result :

f(-1)

Step-by-step explanation:

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