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4 years ago

A 2 cm x 2 cm x 2 cm cube holds 468 grains of rice. How many grains of rice would fit in a box with

1 answer:

4 0

Let the Volume of the first cube be V1

V1 = 2 * 2 * 2

= 8 cm*3

Let the Volume of the second cube be V2

V2 = 6 * 2 * 12

= 144 cm*3

Now, cube1 with volume of 8 cm*3 holds 468 grains of rice.

Thus, 1 cm*3 volume can hold 58.5 grains of rice ( 468 / 8 = 58.5)

Cube2 has volume of 144 cm*3.

Hence, it can hold 8424 grains of rice ( 144 * 58.5 = 8424)

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We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab

777dan777 [17]

Answer:

(a) The distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

Step-by-step explanation:

It is provided that random variables $X_{i}$ are independent and identically distributed Bernoulli random variables with p = 0.50.

The random sample selected is of size, n = 100.

(a)

Theorem:

Let $X_{1},\ X_{2},\ X_{3},...\ X_{n}$ be independent Bernoulli random variables, each with parameter p, then the sum of of thee random variables, $X=X_{1}+X_{2}+X_{3}...+X_{n}$ is a Binomial random variable with parameter n and p.

Thus, the distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

The sample size is large, i.e. n = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu=p=0.50$

And the standard deviation of the distribution of sample mean is given by,

$\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05$

(c)

Compute the value of $P(\bar X>0.50)$ as follows:

$P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\$

$=P(Z>0)\\=1-P(Z$

*Use a z-table.

Thus, the value of $P(\bar X>0.50)$ is 0.50.

8 0

3 years ago

A school director Must randomly select 6 teachers to participate in a training session there are 30 teachers at the school in ho

Ede4ka [16]

Solution:

we are given that

A school director Must randomly select 6 teachers to participate in a training session there are 30 teachers at the schoo. The order of selection does not matter.

As we know

6 teacheras can be selected out of 30 teacher in ${30}_C_6$ ways.

${30}_C_6=\frac{30!}{6!24!}\\
\\ {30}_C_6=\frac{25*26*27*28*29*30}{1*2*3*4*5*6}\\
\\ {30}_C_6=593775$

Hence the required number of ways is 593775.

7 0

3 years ago

2z-18-4z=-8 what is the value of z

zmey [24]

The z is 5

−2⋅z=−8+18

⋅

Simplify :

−2⋅z=10

⋅

Dividing by the variable coefficient :

z=−

Simplify :

z=−5

The solution of equation

2⋅z−18−4⋅z=−8

⋅

[−5]

7 0

3 years ago

The price of Stock A at 9 A.M. was $14.83. Since then, the price has been increasing at the rate of $0.12 each hour. At noon

BARSIC [14]

Answer:

0.5 hours

Step-by-step explanation:

Let the x be the number of hours after noon.

We have the following equation:

14.83 + 0.12(x + 3) = 15.33 - 0.15x
14.83 + 0.12x + 0.36 = 15.33 - 0.15x
0.12x + 0.15x = 15.33 - 15.19
0.27x = 0.14
x = 0.14/0.27
x ≈ 0.5

5 0

3 years ago

Construct a probability distribution for the data.

lianna [129]

Answer:

X : ___ 0 ____ 1 _____ 2 ______ 3

P(X) : _ 6/15 __ 5/15__ 3/15 ____ 1/15

Step-by-step explanation:

From the data, to produce a probability distribution for the data :

X : number of times blood is drawn ;

P(x) : probability that blood is drawn at X times

Hence, the probability distribution table for the data Given goes thus :

X : ___ 0 ____ 1 _____ 2 ______ 3

P(X) : _ 6/15 __ 5/15__ 3/15 ____ 1/15

Probability that blood is drawn 0 times = 6/15

Probability that blood is drawn 1 time = 5/15

Probability that blood is drawn 2 times = 3/15

Probability that blood is drawn 3 times = 1/15

(6/15 + 5/15 + 3/15 + 1/15) = 1

5 0

3 years ago