All categories

3 years ago

Solve 5(x+9) - 4 = 71 for x

Mathematics

2 answers:

MrRissso [65]3 years ago

7 0

Solving for x you would get X=6

Drupady [299]3 years ago

6 0

Answer:

x = 6

Step-by-step explanation:

5 times x is 5x, 5 times 9 is 45; 5x+49-4; 49 minus 4 is 45; 5x+45 = 75; now minus 45 from each side so it would now be 5x = 30; now divide; 30/5 = 6.

You might be interested in

Graph h(x) =(x-2)(x-4)

sveta [45]

Answer:

Check the image below

Step-by-step explanation:

Some Points that May Help: (2,0), (4,0), (1,3), (5,3), (0,8), and (6,8)

4 0

2 years ago

0.001 in Standard form

horrorfan [7]

1*10<span>- 3 is the standard form of 0.001</span>

8 0

3 years ago

What ordered pair is a solution to y = x + 5 and x - 5y = -9?<br> I also need to show the work.

Margarita [4]

Answer:

(- 4, 1 )

Step-by-step explanation:

Given the 2 equations

y = x + 5 → (1)

x - 5y = - 9 → (2)

Substitute y = x + 5 into (2)

x - 5(x + 5) = - 9 ← distribute and simplify left side

x - 5x - 25 = - 9

- 4x - 25 = - 9 ( add 25 to both sides )

- 4x = 16 ( divide both sides by - 4 )

x = - 4

Substitute x = - 4 into either of the 2 equations and evaluate for y

Substituting into (1)

y = - 4 + 5 = 1

Solution is (- 4, 1 )

3 0

2 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

2 years ago

Do this to get brainiest!

IceJOKER [234]

Answer: 8. 3^4+n^4, 9. 7^3+m^4, and 10. s+2t^3.

Step-by-step explanation:

3 0

3 years ago