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1 year ago

Hello can you please help me with one or two please

Mathematics

1 answer:

Zolol [24]1 year ago

6 0

For question number 2, the answer is A 6x+9

Explanation:

$\begin{gathered} 3(2x+3) \\ \text{Clearing the brackets, we get} \\ 6x+9 \end{gathered}$

Question Number 19:

$\begin{gathered} 8x-2x+4 \\ 6x+4 \\ \text{Factorizing, we get} \\ 2(3x+2) \end{gathered}$

So, the correct answer for question number 19 is 6x + 4.

You might be interested in

There are n machines in a factory, each of them has defective rate of 0.01. Some maintainers are hired to help machines working.

frosja888 [35]

Answer:

a) 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!

b) 1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!

c) ∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³

Step-by-step explanation:

Given that;

if n ⇒ ∞

p ⇒ 0

⇒ np = Constant = λ, we can apply poisson approximation

⇒ Here 'p' is small ( p=0.01)

⇒ if (n=large) we can approximate it as prior distribution

⇒ let the number of defective items be d

so p(d) = ((e^-λ) × λ) / d!

NOW

Let there be x number of repairs, So they will repair 20x machines on time. So if the number of defective machine is greater than 20x they can not repair it on time.

λ[n0.01]

p[ d > 20x ] = 1 - [ d ≤ 20x ]

= 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!

Similarly in this case if number of machines d > 80x/3;

Then it can not be repaired in time

p[ d > 80x/3 ]

1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!

n = 300, lets do it for first case i.e;

p [ d > 20x } ≤ 0.01

1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.01

⇒ ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.99

⇒ ∑²⁰ˣ_k=0 (λ^k)/k! = 0.99e^λ

∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³

8 0

2 years ago

Name the line and plane shown in the diagram. A.AB and plane ABD B.BA and plane DC C.AC and plane ABD D.AB and plane DA

elena-14-01-66 [18.8K]

Answer: The line is AB and the plane is ABD, the first option is the correct one.

Step-by-step explanation:

Ok, first some definitions.

A line is any line that crosses two colinear points. Particularly, you can see in the graph that the line crosses through A and B, so the line is AB.

A plane needs 3 non-colinear points (if the points where colinear, then the points may define a line). Other definition of plane is "a line and a point that is not in the line"

So, if our line is AB, then the possible planes are:

ABC and ABD.

then the correct option is:

Line AB and plane ABD, so the correct option is the first one.

6 0

2 years ago

The circumference of a circle is 37. 68 ft. What is the radius of the circle?.

mixer [17]

Answer:

r = 6 ft

Step-by-step explanation:

3 0

2 years ago

Help please I'm to stupid for this

tangare [24]

Answer:

X>0, so the second one

Step-by-step explanation:

Since the point on 0 is OPEN that means the number (0) is NOT part of the solution. Therefore, X is greater than 0.

4 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .