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2 years ago

13

Find The Value Of X. The Diagram Is Not Drawn To Scale.

1 answer:

IgorLugansk [536]2 years ago

3 0

Answer:

evrone keeps asking me to find their x and then I have no time to find 'MY'X

CRIES

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Explain why the polynomial 100x^2 + 150x + 49 is not a perfect square.

Ad libitum [116K]

Answer: The correct answer is Choice C.

For this polynomial to be a perfect square, it would need to be:
(10x + 7)^2

This will ensure that the first terms and the last terms will be 100x^ and 49. However, if you use foil to multiply the factors, you will not get 150x for the center term. Choice C also states that 150x will not be the middle term.

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3 years ago

Can Someone Help Me With This:

bija089 [108]

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2 years ago

Eight times the sum of a number and six

user100 [1]

Answer:

There is any answer as you choose.

Step-by-step explanation:

There is no equal to so you can choose any number.

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3 years ago

Using the completing-the-square method, rewrite f(x) = x2 − 6x + 2 in vertex form.

Irina18 [472]

Answer:

f(x)=(x-3)^2-7

Step-by-step explanation:

Vertex form: f(x) = a(x-h)^2+k

x^2-6x+9+2=f(x)+9

(x-3)^2+2=f(x)+9

f(x)=(x-3)^2-7

8 0

3 years ago

How many randomly chosen guests should I invite to my party so that the probability of having a guest with the same birthday as

dimulka [17.4K]

Answer:

At least 401 chosen guests

Step-by-step explanation:

Represent those with same birth day with Same

So:

$P(Same) > \frac{2}{3}$

Represent number of people with n

There are 365 days in a year and 364 days out of these days is not your birthday.

So, there the probability that n people do not share your birthday is $(\frac{364}{365})^n$

i.e.

$P(none) = (\frac{364}{365})^n$

Solving further, we have that:

$P(same) + P(none) = 1$

$P(same) + (\frac{364}{365})^n = 1$

$P(same) = 1 - (\frac{364}{365})^n$

We're calculating the probability that $P(Same) > \frac{2}{3}$ .

So, we have:

$1 - (\frac{364}{365})^n > \frac{2}{3}$

Collect Like Terms

$1 - \frac{2}{3}> (\frac{364}{365})^n$

$\frac{3-2}{3}> (\frac{364}{365})^n$

$\frac{1}{3}> (\frac{364}{365})^n$

$3^{-1} > (\frac{364}{365})^n$

Take natural logarithm (ln) of both sides

$ln(3^{-1}) > ln((\frac{364}{365})^n)$

$-ln\ 3 > n\ ln\ (\frac{364}{365})$

Apply laws of logarithm

$-ln\ 3 > n\ (ln(364) - ln(365))$

Multiply through by -1

$ln\ 3 < n\ (ln(365) - ln(364))$

Solve for n

$\frac{ln\ 3}{(ln(365) - ln(364))} < n$

Reorder

$n > \frac{ln\ 3}{(ln(365) - ln(364))}$

$n > \frac{1.09861228867}{(5.89989735358 - 5.89715386764)}$

$n> \frac{1.09861228867}{0.00274348594}$

$n > 400.443928891$

$n > 400$ (approximated)

This implies that n = 401, 402, 403 .....

i.e

$n \geq 401$

So, at least 401 people has to be invited

5 0

3 years ago