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3 years ago

Please help it has to be in fraction form rounded all the way

Mathematics

1 answer:

Sever21 [200]3 years ago

3 0

Answer:

The answer to the question provided is

0 = $- \frac{6}{5}$

It has no solutions.

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Can someone PLEASE help me solve this equation ? due soon

RideAnS [48]

$\sf{Given : 3tanx + 7 = \dfrac{2}{(1 - sinx)(1 + sinx)}}$

We know that : (a - b)(a + b) = a² - b²

$\implies \sf{3tanx + 7 = \dfrac{2}{1 - sin^2x}}$

We know that : 1 - sin²x = cos²x

$\implies \sf{3tanx + 7 = \dfrac{2}{cos^2x}}$

$\sf{\bigstar \ \ We \ know \ that : \boxed{\sf{\dfrac{1}{cos^2x} = sec^2x}}}$

$\implies \sf{3tanx + 7 = 2sec^2x}$

We know that : sec²x = 1 + tan²x

$\implies \sf{3tanx + 7 =2(1 + tan^2x)}$

$\implies \sf{2 + 2tan^2x - 3 tanx - 7 = 0}$

$\implies \sf{2tan^2x - 3 tanx - 5 = 0}$

$\implies \sf{2tan^2x - 5tanx + 2tanx - 5 = 0}$

$\implies \sf{2tanx(tanx + 1) - 5(tanx + 1) = 0}$

$\implies \sf{(tanx + 1)(2tanx - 5) = 0}$

$\implies \sf{tanx = -1 \ (or) \ tanx = \dfrac{5}{2} }$

8 0

3 years ago

What is the Factor of 36y^82-4^122

Anuta_ua [19.1K]

the factor for this expression is 7

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3 years ago

Help<br> I dont know the answer im not very smart

Dahasolnce [82]

The domain of G is 4 and the range of G is 9 (domain=x, range=y :))

8 0

3 years ago

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi

svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

3% of the components are identified as defective

This means that $p = 0.03$

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is $P(X = 0)$ when $n = 90$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when $n = 102$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when $n = 105$

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

3 0

3 years ago

4. A fruit vendor packed 100 apples each in 64 baskets. Later, he found some apples were left unpacked. So he packed 30 more app

topjm [15]

Answer:

Here is the answer

Step-by-step explanation:

Apples in 1 baskets=100

Apples in 64 baskets=100*64-6400

apples (before adding 30 apples in each

basket).... ..(1)

Some apples left unpacked

So he added 30 apples to each

Total apples added by him- 30*64(in each basket*no of baskets)

= 1920 apples

.(2)

So total no apples eq(1)+eq(2)

= 6400 apples + 1920 apples

=8320 apples packed in total

Therefore the total no. of apples packed were 8320 apples.

Hope this helps

Thanks

4 0

2 years ago