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3 years ago

I need hele on what I am doing help me

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

8 0

Answer:

ok heres some help

Step-by-step explanation:

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(a + b - c )(a + b + c )

denis-greek [22]

Answer:

$a^2+b^2-c^2+2ab$

Step-by-step explanation:

$(a+b-c)(a+b+c)=a^2+ab+ac+ab+b^2+bc-ac-bc-c^2=a^2+b^2-c^2+2ab$

Hope this helps!

5 0

3 years ago

Read 2 more answers

Find the required measurements of the following trapezoids. b1 = 12 cm b2 = 16 cm h = 9 cm. Compute the area. ? cm2

Nastasia [14]

Area= (b1+b2)/h
= (12+16)/9
= 28/9 = 3.111

5 0

3 years ago

Read 2 more answers

How do you divide thirty one point five by ten?

Romashka-Z-Leto [24]

All divisions by powers of ten entail moving the decimal point left by the number of zeroes. So 31.5/10 = 3.15.

6 0

4 years ago

ann arranges her marbles in groups, with 8 marbles in each group. She writes the fraction 5/8 to show the fraction of marbles in

DerKrebs [107]

If there are 8 marbles in a group , and 5 out of every 8 is red, than in six groups there would be 5, 10, 15, 20, 25, 30 red marbles.

This lists the group of 6 sets.

30 red marbles in 6 sets.

8 0

4 years ago

A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience

bulgar [2K]

Answer:

a) $P(X \geq 1) = 1-P(X$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}$

And replacing we got:

$P(X \geq 1) =1 -1x10^{-18} \approx 1$

b) $P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355$

$P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And adding we got:

$P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992$

c) $P(X \leq 8) =1 -P(X>8) = 1-P(X=9)$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And replacing we got:

$P(X \leq 8)= 1-0.913517=0.086483$

Step-by-step explanation:

Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:

$X \sim Binom(n=9, p=0.99)$

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:

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

Part a

We want to find this probability:

$P(X \geq 1) = 1-P(X$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}$

And replacing we got:

$P(X \geq 1) =1 -1x10^{-18} \approx 1$

Part b

$P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9)$

$P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355$

$P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And adding we got:

$P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992$

Part c

$P(X \leq 8) =1 -P(X>8) = 1-P(X=9)$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And replacing we got:

$P(X \leq 8)= 1-0.913517=0.086483$

4 0

3 years ago