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WINSTONCH [101]
3 years ago
15

Good night, logging out !

Mathematics
1 answer:
musickatia [10]3 years ago
5 0

Good night :). Sleep tight!

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Pls help this so hard
horrorfan [7]

Answer:

616 cm²

Step-by-step explanation:

small cube (4 sides) = 8 x 3 x 4 = 96

large cube (4 sides) = 7 x 10 x 4 = 280

top and bottom of large cube = 12 x 10 x 2 = 240

(we would deduct the 3 x 3 area from  the top of the large cube and add a 3 x 3 area for the top of the small cube. So that cancels each other out.)

total surface area = 96 + 280 + 240 = 616 cm²

6 0
3 years ago
What is the scale factor of the dilation shown?
dmitriy555 [2]
It is 2/3 of the original so I think the answer is 2/3
7 0
3 years ago
(3x^-7 yz^-3)(3x)^-4
Alex

Answer:

The step wise solution is shown below:

Step-by-step explanation:

                (3x^{-7} y z^{-3} ) (3x)^{-4}

            =  (3x^{-7} (3x)^{-4}  y z^{-3} )

            = 3^{1-4} x^{-7-4} y z^{-3}

            = 3^{-3} x^{-11} yz^{-3}

            = \frac{y}{3^{3} x^{11}z^{3}  }

            = \frac{y}{27 x^{11}z^{3}  }

8 0
3 years ago
In ΔBCD, the measure of ∠D=90°, the measure of ∠B=71°, and DB = 67 feet. Find the length of BC to the nearest tenth of a foot.
PSYCHO15rus [73]

Answer:

in triangle BCD

relationship between base and hypotenuse is given by cos angle

cos 71=b/h

cos71=BD/BC

cos71=67/x

x=67/cos71=205.8ft.

the length of BC to the nearest tenth of a foot is 206ft.

8 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
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