All categories

2 years ago

WHAT IS 5/8 PAGE IN 2/3 OF A MINUTE

1 answer:

4 0

3.5 divided 1.25 = 2.8 mph
0.625 divided 0.666 = 0.95 pages a minute. so your answer is 0.95 pages per minute

You might be interested in

Find the length of the third side. If necessary, round to the nearest tenth.

blsea [12.9K]

Answer:

Step-by-step explanation:

a² + b² = c²

5² + b² = 13²

25 + b² = 169

b² = 144

b = 12

Answer: 12

5 0

1 year ago

Read 2 more answers

A discounted amusement park ticket cost $12.95 less than the original price p. Write and solve an equation to find the original

noname [10]

R=percentage of discount
$12.95= (1-\frac{r}{100} )p$
so we get
$\frac{12.95}{(1-\frac{r}{100} )}=p$

6 0

2 years ago

1 point

Serga [27]

Answer:

I do not know

Step-by-step explanation:

6 0

2 years ago

Which is the same as moving the decimal point 3 places to the left in decimal number?

oksano4ka [1.4K]

Multiply it by -1000

7 0

2 years ago

Let X be the number of the cars being repaired at a repair shop. We have the following information:

worty [1.4K]

Answer:

(a) Sample Space

$S = \{0,1,2,3\}$

(b) PMF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

(c) CDF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

$S = \{0,1,2,3\}$

Solving (b): The PMF

From the question, we have:

$P(2) = P(1)$

$P(0) = P(3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$ can be represented as:

$P(1) + P(2) = 0.5[P(0) + P(3)]$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(1) + P(1) = 0.5[P(0) + P(0)]$

$2P(1) = 0.5[2P(0)]$

$2P(1) = P(0)$

$P(0)= 2P(1)$

Also note that:

$P(0) + P(1) + P(2) + P(3) = 1$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(0) + P(1) + P(1) + P(0) = 1$

$2P(1) + 2P(0) = 1$

Substitute $P(0)= 2P(1)$

$2P(1) + 2*2P(1) = 1$

$2P(1) + 4P(1) = 1$

$6P(1) = 1$

Solve for P(1)

$P(1) = \frac{1}{6}$

To calculate others, we have:

$P(2) = P(1)$

$P(2) = P(1) = \frac{1}{6}$

$P(0)= 2P(1)$

$P(0) =2 * \frac{1}{6}$ $P(0) =\frac{1}{3}$

$P(3) = P(0) =\frac{1}{3}$

Hence, the PMF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

See attachment (1) for histogram

Solving (c): The CDF ; F(x)

This is calculated as:

$F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)$

For x = 0;

We have:

$P(X \le 0) = P(0)$

$P(X \le 0) = 1/3$

For x = 1

$P(X \le 1) = P(0) + P(1)$

$P(X \le 1) = 1/3 + 1/6$

$P(X \le 1) = 1/2$

For x = 2

$P(X \le 2) = P(0) + P(1) + P(2)$

$P(X \le 2) = 1/3 + 1/6 + 1/6$

$P(X \le 2) = 2/3$

For x = 3

$P(X \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3$

$P(X \le 3) = 1$

Hence, the CDF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

See attachment (2) for histogram

6 0

3 years ago