All categories

3 years ago

Solve each problem by working backward.

Mathematics

2 answers:

ddd [48]3 years ago

4 0

(x+3)/4=16 is the word part or whatever, now ya gotta solve for x:))

OLEGan [10]3 years ago

3 0

Answer: X= 4/3 I hope this helps:)

Step-by-step explanation:

(3x ). • 4= 16

divide by 4

3x = 4

divide by 3

x= 4/3

You might be interested in

-9x5 + 36x4 + 189x3 in factored form

Lera25 [3.4K]

Answer:

Step-by-step explanation:

-9x^3(x^2 - 4x - 21) =

-9x(x - 7)(x + 3) <===

4 0

3 years ago

Pls help I will not let me add the screenshot so I'm going to try and explain. so there is a rectangle the top measures 10m and

-BARSIC- [3]

Answer:

£23

Explanation:

6x10=60
2x1.5=3
3.14x2^2=12.566
60-3-12.566=44.433
44.43/10=4.43
5 bags
5x4.60=23
=£23

3 0

2 years ago

Imagine each letter of the word ""mathematical"" is written on individual pieces of paper and placed in a bag. Should you pick a

Aleksandr-060686 [28]

Answer:

The question has a missing part and here it the complete question:

Imagine each letter of the word “Mathematical” is written on individual pieces of paper and placed in a bag. Should you pick a random letter from that bag, what is the probability that you pick a vowel?

Answer: 5/12

Step-by-step explanation:

The question asks for the probability of picking a vowel.

From the word 'Mathematical', we have a total of 12 letters and a total of 5 vowels (a,e,a,i,a).

Probability of picking a vowel = Total number of times the event(vowel) occurs/total number of possible outcomes(total letters)

$Probability = \frac{5}{12}$

6 0

2 years ago

Harry’s little sister spilled paint on his factor rainbow for the number 42.

astra-53 [7]

Sorry I have no idea Soz

6 0

2 years ago

Read 2 more answers

The weight of people in a small town in Missouri is known to be normally distributed with a mean of 186 pounds and a standard de

OleMash [197]

Answer:

the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

Step-by-step explanation:

The summary of the given statistical data set are:

Sample Mean = 186

Standard deviation = 29

Maximum capacity 3,417 pounds or 17 persons.

sample size = 17

population mean =3417

The objective is to determine the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds

In order to do that;

Let assume X to be the random variable that follows the normal distribution;

where;

Mean $\mu$ = 186 × 17 = 3162

Standard deviation = $29* \sqrt{17}$

Standard deviation = 119.57

$P(X>3417) = P(\dfrac{X - \mu}{\sigma}>\dfrac{X - \mu}{\sigma})$

$P(X>3417) = P(\dfrac{3417 - \mu}{\sigma}>\dfrac{3417 - 3162}{119.57})$

$P(X>3417) = P(Z>\dfrac{255}{119.57})$

$P(X>3417) = P(Z>2.133)$

$P(X>3417) =1- 0.9834$

$P(X>3417) =0.0166$

Therefore; the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

5 0

3 years ago