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2 years ago

I’ll mark as BRANLIEST!!

Mathematics

2 answers:

ioda2 years ago

5 0

Answer: the answer is C

Liono4ka [1.6K]2 years ago

4 0

Answer:

5n+6 because it says

*6 more than, so it is +6 or 6+

*5 times a number, so it is 5n

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–8(n + 7) >-96??????????????

mihalych1998 [28]

Answer:

D) n<5

Step-by-step explanation:

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2 years ago

9 less than k is 1. Solve the equation

Zinaida [17]

Answer:

the answer is 8

Step-by-step explanation:

9 - 8 = 1

Hope this helps
Brainliest plz

4 0

2 years ago

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Given that ƒ(x) = x² + 4x, evaluate ƒ(-2)

Amiraneli [1.4K]

F(-2)=(-2)2+4(-2)
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3 0

2 years ago

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Check the true statements below:

valentinak56 [21]

Answer:

a) False

b) False

c) True

d) False

e) False

Step-by-step explanation:

a. A single vector by itself is linearly dependent. False

If v = 0 then the only scalar c such that cv = 0 is c = 0. Hence, 1vl is linearly independent. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, only a single zero vector is linearly dependent, while any set consisting of a single nonzero vector is linearly independent.

b. If H= Span{b1,....bp}, then {b1,...bp} is a basis for H. False

A sets forms a basis for vector space, only if it is linearly independent and spans the space. The fact that it is a spanning set alone is not sufficient enough to form a basis.

c. The columns of an invertible n × n matrix form a basis for Rⁿ. True

If a matrix is invertible, then its columns are linearly independent and every row has a pivot element. The columns, can therefore, form a basis for Rⁿ.

d. In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix. False

Row operations can not affect linear dependence among the columns of a matrix.

e. A basis is a spanning set that is as large as possible. False

A basis is not a large spanning set. A basis is the smallest spanning set.

3 0

2 years ago

In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is

Alex777 [14]

Answer:

(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2) The probability at least 5 computers are infected is 0.949.

Step-by-step explanation:

The probability that a computer is defective is, p = 0.40.

(1)

Let X = number of computers to be tested before the 1st defect is found.

Then the random variable $X\sim Geo(p)$ .

The probability function of a Geometric distribution for k failures before the 1st success is:

$P (X = k)=(1-p)^{k}p;\ k=0, 1, 2, 3,...$

Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:

P (X ≥ 5) = 1 - P (X < 5)

= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

$=1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078$

Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2)

Let Y = number of computers infected.

The number of computers in the company is, n = 20.

Then the random variable $Y\sim Bin(20,0.40)$ .

The probability function of a binomial distribution is:

$P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...$

Compute the probability at least 5 computers are infected as follows:

P (Y ≥ 5) = 1 - P (Y < 5)

= 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)] $=1-[{20\choose 0}(0.40)^{0}(1-0.40)^{20-0}+{20\choose 1}(0.40)^{1}(1-0.40)^{20-1}\\+{20\choose 2}(0.40)^{2}(1-0.40)^{20-2}+{20\choose 3}(0.40)^{3}(1-0.40)^{20-3}\\+{20\choose 4}(0.40)^{4}(1-0.40)^{20-4}]\\=1-[0.00004+0.00049+0.00309+0.01235+0.03499]\\=1-0.05096\\=0.94904$

Thus, the probability at least 5 computers are infected is 0.949.

7 0

2 years ago