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2 years ago

Now, solve this problem. 106% of $ _______ = $7.95

Mathematics

1 answer:

Alex787 [66]2 years ago

4 0

Answer:

7.5

Step-by-step explanation:

106% of $x = $7.95

In other words,

$\frac{106}{100}$ of $x = 7.95

Multiplying both sides of the equation by $\frac{100}{106}$

$\frac{106}{100} * \frac{100}{106}$ of x = 7.95 * $[\frac{100}{106}]$

=> x = $[\frac{750}{100}]$

=> x = $[\frac{7.50}{1}]$

=> x = 7.5

Validation: 106% of 7.5 = 7.95

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A customer visiting the suit department of a certain store will purchase a suit with probability .22, a shirt with probability .

BigorU [14]

Answer:

a) The probability that he doesnt but any items is 0.49

b) He buys exactly 1 of those items with probability 0.28

Step-by-step explanation:

lets call su the event that the customer purchases a suit, sh the event that teh customer purchases a shirt and t the event that the customer purchases a tie.

Remembe that for events A, B and C we have that

P(A U B) = P(A) + P(B) - P(A ∩ B)

P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

Also, we are given that

P(su) = 0.22

P(sh) = 0.3

p(t) = 0.28

p(su ∩ sh) = 0.11

P(su ∩ t) = 0.14

P(sh ∩ t) = 0.1

P(sh ∩ t ∩ su) = 0.06

The event that he doesnt buy any item has as complementary event su ∪ sh ∪ t, therefore

P( he doesnt but any items) = 1-P(su U sh U t) =

1-( P(su) + p(sh) + p(t) - P(su ∩ sh) - p(su∩t) - p(sh∩t) + p(su∩sh∩t) ) =

1-(0.22+0.30+0.28-0.11-0.14-0.1+0.06) = 1-0.51 = 0.49

b) The probability that he buys at least 2 items is equal to

p(su ∩ t) + p(su ∩ sh) + p(sh ∩ t) -2 p(su ∩ t ∩ sh) (because we are counting the triple intersection 3 times, so we need to remove it twice)

This number is

0.14+0.11+0.1-2*0.06 = 0.23

Thus, the probability that he buys exactly one item can be computed by substracting from one the probability of the complementary event : she buys 2 or more or non items

P(he buys exactly one item) = 1- ( p(he buys none items) + p(he buys at least 2) ) = 1- 0.49-0.23 = 0.28

7 0

2 years ago

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On a road map, the locations A, B and C are collinear. Location C divides the road from location A to B, such that AC:CB = 1:2.

koban [17]

Answer:

C. (-1,-2)

Step-by-step explanation:

Since C internally divides AB in the ratio AC/CB = 1/2 = m/n where m = 1 and n = 2, we use the formula for internal division.

Let A = (x₁, y₁) = (5, 16), B = (x₂, y₂) and C = (x, y) = (3, 10)

So x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Substituting the values of the coordinates, we have

x = (mx₂ + nx₁)/(m + n)

3 = (1 × x₂ + 2 × 5)/(2 + 1)

3 = (x₂ + 10)/3

multiplying through by 3, we have

9 = x₂ + 10

x₂ = 9 - 10

x₂ = -1

y = (my₂ + ny₁)/(m + n)

10 = (1 × y₂ + 2 × 16)/(2 + 1)

10 = (x₂ + 32)/3

multiplying through by 3, we have

30 = y₂ + 32

y₂ = 30 - 32

y₂ = -2

So, the coordinates of B are (-1, -2)

3 0

1 year ago

Write the definition of the vertex of an angle as a biconditional statement

daser333 [38]

The highest point,the top, or apex it is just like a mountain,its vertex is called an peak

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2 years ago

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Paul is going to have the exterior of his home painted. Painting USA charges $75.00 to come out and evaluate the house plus $25.

murzikaleks [220]

CP - the cost of painting the house for Painting USA company [$]
cP = 75 + 25 * 25 = 75 + 625 = 700 [$]
cU - the cost of painting the house for Upscale Home Painting company [$]
cU = 0 + 40 * 25 = 1000 [$]

cP < cU
The most cost effective is Painting USA company.

n - number of houres required to paint the house if the same cost for both companies
75 + 25 * n = 40 * n
40 n - 25n = 75
15n = 75 /:15
n = 5
The cost of paining the house is the same for both companies if the number of houres required to paint the house is 5.

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2 years ago

The probability of flowering plants bearing flowers in January is given the table. If a plant has flowered in January, what is t

ololo11 [35]

Answer:

Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

Step-by-step explanation:

We know that:

Probability =Total favourable outcomes/ Total possible outcomes

Probability of flowering plants bearing flowers in January is given in the table. It means that out of 100 peony 70 peony flowers in January.

There are five kinds of flowers.

We have to select peony and find it's Probability.

As all flowers are independent, their flowering phenomenon does not depend on each other.

So,Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

7 0

1 year ago

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