All categories

3 years ago

What are the roots of the equation?

2 answers:

7 0

For the equation to = 0 , one of the components must = 0
x = 0
x = -13/9
x = -2/11

(Make each bracket = 0)
e.g. 9x+13=0
9x = -13
x = -13/9

quester [9]3 years ago

7 0

Answer:

x = 0

x = -13/9

x = -2/11

Step-by-step explanation:

You might be interested in

Volume of a hemisphere formula?

Allisa [31]

To find the volume of a sphere you use...

V = 4/3 pi r^3

"Hemi-" means half. So we would divide by 2.

V= (4/3 pi r^3) / 2

8 0

3 years ago

Evaluate h(5t^{2})given that h(t)=17-11t

Gala2k [10]

$h(t)=17-11t$
$h(5t^{2})=17-11(5t^{2})$
$h(5t^{2})=17-55t^{2}$

5 0

4 years ago

Five times the sum of three times a number and 4 is -10, find the number.

Minchanka [31]

Answer:

-2

Step-by-step explanation:

5[3x+4)]=-10

15x+20= -10

15x=-30

x= -2

5 0

4 years ago

A New York Fire company is installing safety ladders to an apartment complex. Ladder A is 26 feet long and is placed so the base

Rina8888 [55]

Ladder A because B is shorter in length and is also further away from the building, making Ladder A the highest vertical distance.

4 0

3 years ago

) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.

yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the 0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= $\dfrac{1}{2}$

The length of bits : n = 10

Let X = Number of getting ones.

Then , $X \sim Bin(n=10,\ p=\dfrac{1}{2})$

Binomial distribution formula : $P(X=x)=^nC_x p^x q^{n-x}$ , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , $p=q=\dfrac{1}{2}$

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

$P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}$

$=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}$

$=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})$

$=(\dfrac{1}{2})^{10}(210+252)$

$=(0.0009765625)(462)$

$=0.451171875\approx0.4512$

Hence, the probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0

4 years ago