Can someone please help me with this question?<br><br> Find the value of c-5 when c=14

What is the volume of a gift box in the shape of a rectangular prism that is 3.5 inches high, 7 inches long, and 6 inches wide

ankoles [38]

Answer:

147 cubic inches

Step-by-step explanation:

volume of the prism is length x breadth x height

3.5 x 7 x 6=147

8 0

2 years ago

Could someone please help me with this problem and possibly explain if you can? (Will give up brainiest answer if correct, if po

Andrej [43]

To find the answer you need to find the square root of e^2 to get e so what you do to one side of the equation you do to the other side of the equation so find the square root of 0.36.
e=positive or negative 0.6

3 0

3 years ago

This tree diagram shows the possible outcomes for tossing a nickel and tossing a penny. What is the probability of both coins la

Elanso [62]

Answer:

The possibility is 1/4, with 1 representing both coins tails up, and the 4 representing all the outcomes possible.

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

2 years ago

Find the x and y intercepts

UNO [17]

Answer:

2x

Step-by-step explanation:

solve it

5 0

3 years ago

Can someone please help me with this question? Find the value of c-5 when c=14