PLEASE HELP ME OUT! i need to get this!

The smiths want to fence their square garden, which has an area 256 square meters.

diamong [38]

The answer s=16 answer 48

6 0

3 years ago

Nan’s art sculpture is a pyramid. She will cover all the faces with gems, except the base of the pyramid.

marysya [2.9K]

Answer:

it is 54

Step-by-step explanation:

i took the test and checked the answer

6 0

3 years ago

Read 2 more answers

I need help on this problem. Does anyone know how to complete it? Please and thank you :)

ale4655 [162]

I think it's 5 because i remember learning that it's gonna be half of what the bottom is

4 0

3 years ago

According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e

slamgirl [31]

Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that $p = 0.09$

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

So

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225$

$P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827$

$P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082$

$P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982$

Finally

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180$

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

6 0

3 years ago

Solve for r and t:<br> 6r - 3t = 6<br> 6r + 8t = -16

MAVERICK [17]

Multiply the first equation with (-1)
you should get this

-6r + 3t = -6
6r + 8t = -16

start to eliminate

11t = -22

divide by 11

t = -2

substitute

6r - 3(-2) = 6
6r + 6 = 6
6r = 6 - 6
6r = 0
r=0

8 0

3 years ago