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3 years ago

Find the density of an object that has a volume of 20 cm3 and a mass of 56g

Mathematics

1 answer:

kherson [118]3 years ago

7 0

Answer: 34y

Step-by-step explanation:

So first you have to round the nearest 2 to the nears 3 then you get 30 then you have to add 4 the add you variable.

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Suppose s varies directly with t. When t=27, s=3. What is s when t=63?

adell [148]

Answer: s = 7

Step-by-step explanation:

27 is simply 9 times 3.

Thus, simply do 63/9 to get 7.

Hope it helps :)

3 0

3 years ago

Read 2 more answers

The ratio of the measures of the sides of a triangle is 8: 15: 17, and its perimeter is 480 in.. Find the measure of the shortes

lubasha [3.4K]

Answer:

Step-by-step explanation:

The ratio of the measure of sides of a triangle are 8:15:17

The perimeter is 480

8x + 15x + 17x= 480

40x= 480

x = 480/40

x = 12

Therefore the measure of the shortest side of the triangle can be calculated as follows

8×12

= 96

6 0

3 years ago

What is the next term for the given arithmetic sequence? -3, -2.25, -1.5, -0.75, ...

Rasek [7]

0. Each time you're adding 0.75, so adding $-0.75+0.75=0$ !

5 0

3 years ago

Read 2 more answers

Please help

nalin [4]

Answer:

1,2,3,4 on edge

Step-by-step explanation:

8 0

2 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago