What is 0.08333333333 as a fraction?

Four subtracted from eight is equal to two squared

Scilla [17]

Yes. This is true because 4 subtracted from 8 is 8-4=4. The value you get from subtracting 4 from 8 equals the same value you would get if you were to take the value of 2 and square it. Both equal 4.

4 0

3 years ago

Honda Atlas issued 100,000 shares of stock. In 1990, each share of stock was worth Rs. 152.50. In 1993, each share of the stock

noname [10]

Answer:

Rs.13,130 ,000

Step-by-step explanation:

Lets solve first for year 1990

Value of stock in 1990 = Rs. 152.50

No. of stocks issued in 1990 = 100,000

Total value of these stocks in 1990= Value of stock in 1990* no of stocks issued in 1990 = 152.50*100,000 = Rs. 15,250,000

Let solve first for year 1993

Value of stock in 1993 = Rs. 21.20

No. of stocks 1993 = 100,000

Total value of these stocks in 1993 = Value of stock in 1993* no of stocks issued in 1993 = 21.20*100,000 = Rs. 2,120,000

Decrease in value of stock in 1993 in comparison to 1990 = Total value of the stocks in 1990 -Total value of the stocks in 1993

= Rs. 15,250,000 - Rs. 2,120,000

= Rs.13,130 ,000

Thus, the 100,000 shares worth Rs.13,130 ,000 lesser in 1993 than in 1990.

6 0

3 years ago

The mean amount spent by a family of four on food per month is $500 with a standard deviation of $75. Assuming that the food cos

JulijaS [17]

Answer:

The probability that a family spends less than $410 per month

P( X < 410) = 0.1151

Step-by-step explanation:

Step(i):-

Given mean of the population = 500

Given standard deviation of the Population = 75

Let 'X' be the variable in normal distribution

$Z = \frac{x-mean}{S.D}$

Given X = $410

$Z = \frac{410-500}{75} = - 1.2$

Step(ii):-

The probability that a family spends less than $410 per month

P( X < 410) = P( Z < - 1.2 )

= 0.5 - A( -1.2)

= 0.5 - A(1.2)

= 0.5 - 0.3849 ( ∵from normal table)

= 0.1151

Final answer:-

The probability that a family spends less than $410 per month

P( X < 410) = 0.1151

6 0

2 years ago

Evaluate −x^2−5 y^3 when x = 4 and y = 1

Natali [406]

Answer:

Simplify:

$-4^2-5(1^3)$

So you get:

$-21\\$

4 0

2 years ago

Read 2 more answers

Four teams A,B,C, and D compete in a tournament, and exactly one of them will win the tournament. Teams A and B have the same ch

beks73 [17]

Answer:

A= 0,2

B= 0,2

C= 0,4

D=0,2

Step-by-step explanation:

We know that only one team can win, so the sum of each probability of wining is one

P(A)+P(B)+P(C)+P(D)=1

then we Know that the probability of Team A and B are the same, so

P(A)=P(B)

And that the the probability that either team A or team C wins the tournament is 0.6, so P(A)+Pc)= 0,6, then P(C)= 0.6-P(A)

Also, we know that team C is twice as likely to win the tournament as team D, so P(C)= 2 P(D) so P(D) = P(C)/2= (0.6-P(A))/2

Now if we use the first formula:

P(A)+P(B)+P(C)+P(D)=1

P(A)+P(A)+0.6-P(A)+(0.6-P(A))/2=1

0,5 P(A)+0.9=1

0,5 P(A)= 0,1

P(A)= 0,2

P(B)= 0,2

P(C)=0,4

P(D)=0,2

4 0

2 years ago