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3 years ago

Help my sister please

Mathematics

1 answer:

Margaret [11]3 years ago

6 0

Answer:

Step-by-step explanation:

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The total sales at Office Products this year are $713,340, which is 35% more than last year’s sales. What is the amount of last

antoniya [11.8K]

The amount of last year sales is $ 528400

Solution:

Let "x" be the amount of last year sales

Given that,

The total sales at Office Products this year are $713,340, which is 35% more than last year’s sales

Therefore,

This year sales = 35 % more than last year sales

This year sales = 35 % of last year sales + last year sales

713340 = 35 % of x + x

$713340 = 35 \% \times x + x\\\\713340 = \frac{35}{100} \times x + x\\\\713340 = 0.35x + x\\\\713340 = 1.35x\\\\\text{Divide both sides of equation by 1.35 }\\\\x = 528400$

Thus amount of last year sales is $ 528400

8 0

3 years ago

a rectangular field has a length that is two meters more than the breadth,if the total area of the field is 168m², calculate the

sdas [7]

Answer:765

Step-by-step explanation:

Jjjjjjh kingdom

6 0

2 years ago

Which of the following is equivalent to 249 ?

Burka [1]

Answer:

Step-by-step explanation:

i^49= i

i=√-1 or i

6 0

3 years ago

a packe of cards conatins 4 red and 5 black cards. a hand of 5 cards is drawn without replacment. What is the proablity of all f

Vladimir79 [104]

Answer: Our required probability is 0.3387.

Step-by-step explanation:

Since we have given that

Number of red cards = 4

Number of black cards = 5

Number of cards drawn = 5

We need to find the probability of getting exactly three black cards.

Probability of getting a black card = $\dfrac{5}{9}$

Probability of getting a red card = $\dfrac{4}{9}$

So, using "Binomial distribution", let X be the number of black cards:

$P(X=3)=^5C_3(\dfrac{5}{9})^3(\dfrac{4}{9})^2\\\\P(X=3)=0.3387$

Hence, our required probability is 0.3387.

5 0

3 years ago

1)cot a/2 -tan a/2 = 2 cot a 2) cot b/2 + tan b/2= 2 cosec b prove

VladimirAG [237]

1)

LHS = cot(a/2) - tan(a/2)

= (1 - tan^2(a/2))/tan(a/2)

= (2-sec^2(a/2))/tan(a/2)

= 2cot(a/2) - cosec(a/2)sec(a/2)

= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))

= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)

= 2[(1+cos(a) -1)/sin(a)]

=2cot a

= RHS

2.

LHS = cot(b/2) + tan(b/2)

= [1 + tan^2(b/2)]/tan(b/2)

= sec^2(b/2)/tan(b/2)

= 1/sin(b/2)cos(b/2)

using product to sums

= 2/sin(b)

= 2cosec(b)

= RHS

8 0

3 years ago