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2 years ago

7-12, in 7&8 trace line m and point m

Mathematics

1 answer:

yKpoI14uk [10]2 years ago

5 0

Answer:

sorry i neeeeeeeeeeeeeed the points

Step-by-step explanation:

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Solve for x: geometry question plz help!!!

hoa [83]

Answer:

Step-by-step explanation:

But not too sure tho make sure of it

8 0

3 years ago

5 10 — = — s 14 Help!

allsm [11]

The answer to your question is 514

7 0

3 years ago

Determine whether the function f(x) = -9.5 + 6 + x² is even, odd or neither.

zimovet [89]

Answer:

The function f(x) = -9.5 + 6 + x² is neither odd or even.

Step-by-step explanation:

We know that a function is termed as 'even' when

f(-x) = f(x) for all x

We know that a function is termed as 'odd' when

f(-x) = -f(x) for all x

Given the function

f(x) = -9.5x⁵ + 6 + x²

substitute x with -x

f(-x) = -9.5(-x)⁵ + 6 + (-x)²

as (-x)⁵ = -x⁵, so

f(-x) = -(-9.5x)⁵ + 6 + (-x)²

Apply exponent rule: (-a)ⁿ = aⁿ, if n is even

f(-x) = -(-9.5x)⁵ + 6 + x²

Apply rule: -(-a) = a

f(-x) = 9x⁵ + 6 + x²

f(-x) ≠ f(x) ≠ -f(x)

Therefore, the function f(x) = -9.5 + 6 + x² is neither odd or even.

5 0

2 years ago

What is the answer for question 7?

Norma-Jean [14]

We know the cofunction identity $tan(90-x)=cotx$ .

Thus, the tangent of 90° minus θ also equals 5.

5 0

3 years ago

The Magazine Mass Marketing Company has received 1515 entries in its latest sweepstakes. They know that the probability of recei

wariber [46]

Answer:

$P(x < 5)=P(X=0)+P(X=1)+P(X=2)+P(x=3)+P(X=4)$

$P(X=0)=(15C0)(0.5)^0 (1-0.5)^{15-0}=0.0000305$

$P(X=1)=(15C1)(0.5)^1 (1-0.5)^{15-1}=0.000458$

$P(X=2)=(15C2)(0.5)^2 (1-0.5)^{15-2}=0.0032$

$P(X=3)=(15C3)(0.5)^3 (1-0.5)^{15-3}=0.0139$

$P(X=4)=(15C4)(0.5)^4 (1-0.5)^{15-4}=0.0417$

And adding we got:

$P(x < 5)=0.0592$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=15, p=0.5)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(x < 5)=P(X=0)+P(X=1)+P(X=2)+P(x=3)+P(X=4)$

$P(X=0)=(15C0)(0.5)^0 (1-0.5)^{15-0}=0.0000305$

$P(X=1)=(15C1)(0.5)^1 (1-0.5)^{15-1}=0.000458$

$P(X=2)=(15C2)(0.5)^2 (1-0.5)^{15-2}=0.0032$

$P(X=3)=(15C3)(0.5)^3 (1-0.5)^{15-3}=0.0139$

$P(X=4)=(15C4)(0.5)^4 (1-0.5)^{15-4}=0.0417$

And adding we got:

$P(x < 5)=0.0592$

7 0

3 years ago