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3 years ago

HURRY ANSWER THIS!

Mathematics

2 answers:

Eddi Din [679]3 years ago

8 0

His brain? Pfft I have no clue honestly but it’s worth a try aha

Ann [662]3 years ago

7 0

Answer:

none lul

Step-by-step explanation:

my noggin

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What type of triangle is this?

olga nikolaevna [1]

This is a equilateral triangle because all the sides are equal.

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3 years ago

Shane and Abha earned a team badge that required their team to collect no less than 2000 cans for recycling. Abha collected 178

dangina [55]

Answer:

$2S+178\ge 2,000,$

$[911,\infty).$

Step-by-step explanation:

Let S be the number of cans that Shane had collected.

Abha had collected 178 more cans than Shane did, then Abha had collected S+178 cans.

Shane and Abha earned a team badge that required their team to collect no less than 2000 cans for recycling, this means that

$S+S+178\ge 2,000,\\ \\2S+178\ge 2,000.$

Solve this inequality:

1. Divide it by 2:

$S+89\ge 1,000.$

2. Now

$S\ge 1,000-89,\\ \\S\ge 911.$

The solution set is $[911,\infty).$

6 0

3 years ago

18-(4x+3y) if x = -5 and y = 3

zaharov [31]

Answer:

this is your answer

I hope it helps you

,:-))

5 0

2 years ago

Roger has 47 cars. Can he group the cars in more than 2 ways?

Mazyrski [523]

Yes he can, he could do:

1 : 46
2 : 45
3 : 44 and so on

7 0

3 years ago

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

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".

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

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

3 years ago