Evaluate: [7(5 – 2) + 42] ÷ 9

PLZ HELP

Mashutka [201]

(5-5) x
It is different because when you distribute the x it will make it 5x-5x while all the other equations are 5-5x

7 0

3 years ago

Urgent please ans this

Rama09 [41]

Answer:

x=6

Step-by-step explanation:

Remember, the formula for area is the length times the width. So, if we know the width is 3, and we know the total area is 24, all we need to do is divide 24 by 3 to get 8. 8 is the split up side, the one that is x + 2. To find x, simply subtract 2 from 8. 8 - 2=6.

24/3=8

8-2=6

3 0

4 years ago

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

a) The probability that exactly 4 flights are on time is equal to 0.0313

b) The probability that at most 3 flights are on time is equal to 0.0293

c) The probability that at least 8 flights are on time is equal to 0.00586

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures.

a) The probability that exactly 4 flights are on time is =

b) The probability that at most 3 flights are on time is =

c)The probability that at least 8 flights are on time is =

<h2>Solution to the problem</h2>

a) Probability that exactly 4 flights are on time

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

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

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

b) Probability that at most 3 flights are on time

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

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

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

c) Probability that at least 8 flights are on time 

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

6 0

4 years ago

What is the derivative of g(x)=e^(x^2+2x)+3x

UNO [17]

Ok first we can split it in two : $e^{x^2+2x}$ and $3x$ .

The derivative of $3x$ is 3.

For the first part, we use the chain rule : $[f(g(x))]'=g'(x)f'(g(x))$ hence $(e^{x^2+2x})'=(x^2+2x)'e^{x^2+2x}$ (since the derivative of the exponential is itself) hence $g'(x)=(2x+2)e^{x^2+2x}+3$

7 0

3 years ago

Read 2 more answers

naomi needs 2 cups of chopped apples for a fruit salad she is making .she only has a 1/4 cup. how many times will naomi need to

Lilit [14]

1/4 × 8 times = 2 cups
ANSWER : 8 times

5 0

3 years ago

Read 2 more answers