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3 years ago

Which answer is right?

Mathematics

2 answers:

jek_recluse [69]3 years ago

6 0

The answer is A

i believe

yanalaym [24]3 years ago

5 0

Answer:

i think A

Step-by-step explanation:

You might be interested in

D ÷ 3 = 138 what is d

V125BC [204]

Answer:

D equals 414

Step-by-step explanation:

414 is correct because to check do 414 divided by 3 to equal 138

MARK ME AS BRANLIEST

4 0

3 years ago

Read 2 more answers

Federal Rent-a-Car is putting together a new fleet. It is considering package offers from three car manufacturers. Fred Motors i

Viefleur [7K]

Answer:

40 packages from Fred Motors
20 packages from Admiral Motors
40 packages from Chrysalis

Step-by-step explanation:

I would formulate the problem like this. Let f, a, c represent the numbers of packages bought from Fred Motors, Admiral Motors, and Chrysalis, respectively. Then the function to minimize (in thousands) is …

objective = 500f +400a +300c

The constraints on the numbers of cars purchased are …

5f +5a +10c >= 700

5f +10a +5c >= 600

10f +5a +5c >= 700

Along with the usual f >=0, a>=0, c>=0. Of course, we want all these variables to be integers.

Any number of solvers are available in the Internet for systems like this. Shown in the attachments are the input and output of one of them.

The optimal purchase appears to be …

40 packages from Fred Motors
20 packages from Admiral Motors
40 packages from Chrysalis

The total cost of these is $40 million.

8 0

4 years ago

Ralph bought a dozen widgets and it cost $30. how much will 80 widgets cost ?

Lesechka [4]

A dozen = 12

12 widgets = $30
80 widgets = $x

$12x = 30 \times 80 \\ 12x = 2400 \\ x = 200$

Answer : 80 widgets would cost $200.

Hope this helps. - M

6 0

3 years ago

A bookstore classifies its books by reader group, type of book, and cost. What is the probability that a book selected at random

Murrr4er [49]

Answer:

Pr(Cb) & Pr(cost $15) = 15(Ncb)/(T1×T3)

Step-by-step explanation:

Classification of the book in to 3

First let's assign terms to each probability

Let:

T1= total number of type of Reader's group

T2 = total number of Type of book

T3 = total number of type of Cost

Probability of child's book and cost $15= Pr(Cb) & Pr(cost $15)

Then find probability of child's book and cost separately:

Pr of child's book = Pr(Cb)

Pr(Cb) = (number of child's book)/ total number of type of group

Ncb = number of child's book

Pr(Cb) = Ncb/T1

Pr (cost $15) = $15 book/total number of cost

Pr (cost $15) = 15/T3

probability of child's book and cost $15 = Pr(Cb) × Pr (cost $15)

= Ncb/T1 × 15/T3

Pr(Cb) & Pr(cost $15) = 15(Ncb)/(T1×T3)

3 0

3 years ago

If a distribution for a quantitative variable is thought to be nearly symmetric with very little variation,and a box and whisker

marusya05 [52]

Answer:

B). The left and right-hand edges of the box will be approximately equal distances from the median.

Step-by-step explanation:

The 'symmetry' si described as a 'satisfying arrangement of a balanced distribution of the elements of the whole' while a 'symmetry group' is characterized as a group whose elements are all the transformations under which a given object remains invariant and whose group operation is function composition.

As per the given conditions, the second statement asserts a true claim regarding the keeping of edges of left, as well as, right-hand side's boxes at equal intervals from the median. This will help in making the arrangement fulfilling while keeping a little scope for the variation. Thus, option B is the correct answer.

6 0

3 years ago