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3 years ago

Need Help Please I'm Out Of Time ;(

Mathematics

1 answer:

STatiana [176]3 years ago

5 0

Answer:

last three questions are statistical question

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If x^2=20 what is the value of x

tankabanditka [31]

Answer:

Step-by-step explanation:

x^2=20

just square both sides and u get

x = 4.472135955

6 0

4 years ago

Read 2 more answers

Tonya has 39 packages. Each package weighs 6 pounds. Choose the best estimate of the total weight of the packages.

Sidana [21]

Step-by-step explanation:

If were estimating, 39 is pretty close to 40 and 40x6 is 240 pounds. if we are getting exact about it, then 234 pounds

5 0

3 years ago

Find an antiderivative F(x) with F′(x) = f(x) = 6 + 24x^3 + 18x^5 and F(1)=0.

7nadin3 [17]

Answer:

The antiderivative is $F(X) = 6x + 6x^4 + 3x^6 - 15$ .

Step-by-step explanation:

Antiderivative F(x)

This is the integral of $F^{\prime}(x)$

F′(x) = f(x) = 6 + 24x^3 + 18x^5

Then:

$F(x) = \int (6 + 24x^3 + 18x^5) dx$

$F(x) = 6x + \frac{24x^4}{4} + \frac{18x^6}{6} + K$

$F(x) = 6x + 6x^4 + 3x^6 + K$

F(1)=0

$F(X) = 0$ when $x = 1$ . We use this to find K.

$F(x) = 6x + 6x^4 + 3x^6 + K$

$0 = 6 + 6 + 3 + K$

$K = -15$

Thus

The antiderivative is $F(X) = 6x + 6x^4 + 3x^6 - 15$ .

7 0

3 years ago

BRIANLIST ANSWER!! Please help on this one question ( if you’re good at this) thanks :))

ivann1987 [24]

Answer:

Step-by-step explanation:

(21²+10²)½ = 23.25= 23.3

5 0

3 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

3 years ago