7 2 − 140 degrees for angles

Expreess 3.81 as a mixed number

Luba_88 [7]

Answer:

3.81 = 3 81/100

Step-by-step explanation:

This can’t be simplified so it stays as 3 $\frac{81}{100}$

8 0

3 years ago

Given a leading coefficient of 8, polynomial roots of 1 & 2, and the known point on the graph (4,5). Write an equation that

m_a_m_a [10]

Given:

The leading coefficient of a polynomial is 8.

Polynomial roots are 1 and 2.

The graph passes through the point (4,5).

To find:

The 3rd root and the equation of the polynomial.

Solution:

The factor form of a polynomial is:

$y=a(x-c_1)(x-c_2)...(x-c_n)$

Where, a is a constant and $c_1,c_2,...,c_n$ are the roots of the polynomial.

Polynomial roots are 1 and 2. So, $(x-1)$ and $(x-2)$ are the factors of the polynomial.

Let the third root of the polynomial by c, then $(x-c)$ is a factor of the polynomial.

The leading coefficient of a polynomial is 8. So, a=8 and the equation of the polynomial is:

$y=8(x-1)(x-2)(x-c)$

The graph passes through the point (4,5). Putting $x=4,y=5$ , we get

$5=8(4-1)(4-2)(4-c)$

$5=8(3)(2)(4-c)$

$5=48(4-c)$

Divide both sides by 48.

$\dfrac{5}{48}=4-c$

$c=4-\dfrac{5}{48}$

$c=\dfrac{192-5}{48}$

$c=\dfrac{187}{48}$

Therefore, the 3rd root on the polynomial is $\dfrac{187}{48}$ .

8 0

2 years ago

Suppose a bag of marbles has 4 green, 2 red, 5 yellow, 1 brown, and 7 blue marbles. What is the probability of picking a red mar

swat32

Assuming that each marble can be picked with equal probability, we notice that there is a total of

$4+2+5+1+7 = 19$

marbles, of which 2 are red.

So, the probability of picking a red marble is

$\dfrac{2}{19}$

In fact, as in any other case of (finite) equidistribution, we used the formula

$P(\text{event}) = \dfrac{\text{number of favourable cases}}{\text{number of all possible cases}}$

4 0

2 years ago

Approximately 12.6% of all (untreated) Jonathan apples had bitter pit in a study conducted by the botanists Ratkowsky and Martin

bagirrra123 [75]

Answer:

Step-by-step explanation:

Let n be a random variable that represents the first Jonathan apple chosen at random that has bitter pit.

a) P(X = n) = q(n-1)p, where q = 1 - p.

From the information given, probability if success, p = 12.6/100 = 0.126

b) for n = 3, the probability value from the geometric probability distribution calculator is

P(n = 3) = 0.096

For n = 5, the probability value from the geometric probability distribution calculator is

P(n = 5) = 0.074

For n = 12, the probability value from the geometric probability distribution calculator is

P(n = 12) = 0.8

c) For n ≥ 5, the probability value from the geometric probability distribution calculator is

P(n ≥ 5) = 0.58

d) the expected number of apples that must be examined to find the first one with bitter pit is the mean.

Mean = 1/p

Mean = 1/0.126 = 7.9

Approximately 8 apples

7 0

3 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

5 0

3 years ago