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4 years ago

Alvin is dividing 0.4 by 0.9.

1 answer:

7 0

Its C, It can't be equal because it is not dividing by 1, It can't be less than 0.4 because if you divide by a decimal the quotient will increase. So the obvious choice is C because it will be greater than 0.4. Also a question for you, why use this website to get answers when there are online calculators....

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Rene is going to the lake to visit some friends if the lake is 60 miles away, and rene is driving a 40 miles per hour the entire

Ostrovityanka [42]

40/60=60/x...forty miles in sixty minutes is equal to 60 miles in 40 minutes.....40x=3600....x=90..90 minutes

4 0

4 years ago

Help! Idk NOT know this will give crown

gregori [183]

Answer:

Step-by-step explanation:

the triangle with the missing lenght is being multiplied by 8. 5 times 8 is 40. have a good day

6 0

3 years ago

How do you find a if given c and 0 (angle) in a right triangle where a is the opposite, c is the hypotenuse, b is adjacent and 0

Yakvenalex [24]

C can be found using pythagoras theorem. c2=a2+b2. Now, b is not given, but we know that cos(theta)=b/c=>b=c*cos(theta). Substituting b in the above relation, c2=a2+c2(cos(theta))^2=>c2=a2/(1-cos((theta))^2). c is the squareroot of c2. Hence c=sqrt(2/(1-cos((theta))^2))

4 0

3 years ago

3^2 x 8<br> A. 78<br> B. 72<br> C. 73<br> D. 84

Fiesta28 [93]

Answer:

Step-by-step explanation:

3^2= 9

9x8 = 72

4 0

3 years ago

According to government data, 44% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women

faltersainse [42]

Solution :

Given :

P (never married) = P = 0.44

Sample size, n = 15

Let $X \sim \text{Binomial}$ (n = 15, P = 0.44)

P(X = x) = $^nC_xP^x(1-P)^{n-x}; x=0, 1,....n$

a). The probability that exactly two of them have been married is

P(X=2) = $^{15}C_2(0.44)^2(1-0.44)^{15-2}$

$=\frac{15!}{2!(15-2)!}(0.44)^2(0.56)^{13}$

$=\frac{15\times14}{2}\times (0.44)^2(0.56)^{13}$

= 0.0108

b). That at most two of them have never been married.

$P(X \leq 2)= P(X=0)+P(X+1)+P(X=2)$

$=^{15}C_0(0.44)^0(1-044)^{15-0}+^{15}C_1(0.44)^1(1-0.44)^{15-1}+^{15}C_2(0.44)^2(1-0.44)^{15-2}$

$=(1)(1)(0.56)^{15}+(15)(0.44)(0.56)^{14}+(105)(0.44)^2(0.56)^{13}$

= 0.000167+0.001969+0.010828

= 0.012964

= 0.0130

c). That at least 13 of them have been married.

P(married) = 1 - P(never married)

= 1 - 0.44

= 0.56

$P(X \geq 13)= P(X=13)+P(X=14)+P(X+15)$

$=^{15}C_{13}(0.56)^{13}(1-0.56)^{15-13}+^{15}C_{14}(0.56)^{14}(1-0.56)^{15-14}+^{15}C_{15}(0.56)^{15}(1-0.56)^{15-15}$ $=(105)(0.56)^{13}(0.44)^{2}+(15)(0.56)^{14}(0.44)+(1)(0.56)^{15}(1)$

$=0.010828+0.001969+0.000167$

= 0.012964

=0.0130

7 0

3 years ago