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3 years ago

If 36^12-m=6^2m, what is the value of m? 4,6,8,9

Mathematics

2 answers:

Viefleur [7K]3 years ago

4 0

The Answer is B.) 6
I Just took the test as well.

stealth61 [152]3 years ago

3 0

$36^{12-m}=6^{2m}\\\\(6^2)^{12-m}=6^{2m}\\\\6^{2(12-m)}=6^{2m}\\\\6^{24-2m}=6^{2m}\iff24-2m=2m\ \ \ |add\ 2m\ to\ both\ sides\\\\4m=24\ \ \ \ |divide\ both\ sides\ by\ 4\\\\\boxed{m=6}$

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Un agricultor a adus la piață 18 saci cu câte 42 de kg de cartofi și 28 de saci cu câte 36 kg de vinete Ce cantitate de legume a

NikAS [45]

Answer:

the number of vegetables did farmer bring in the market is 1764 vegetables

Step-by-step explanation:

The computation of the number of vegetables did farmer bring in the market is shown below:

= 18 bags × 42 kgs + 28 bags × 36 kgs

= 756 + 1008

= 1764 vegetables

Hence, the number of vegetables did farmer bring in the market is 1764 vegetables

6 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

What is the value of x in the circle on the right

german

To solve this we need to use Pythagorean theorem

${x}^{2} + {14}^{2} = (9 + x) ^{2} \\ {x}^{2} + 196 = 81 + 18x + {x}^{2} \\ 18x = 115 \\ x = 6.3(8) \simeq6.4$

Answer: x≈6.4

8 0

3 years ago

Daisy filled a bucket with 7/6 gallon of rain.water.

MissTica

A gallon or 6/6 gallon

5 0

2 years ago

o persoana depune la banca o suma de bani. banca ofera o dobanda anuala de 2%. dupa un an in cont se afla 8160 lei. Care a fost

Lorico [155]

Answer:

8000 lei

Step-by-step explanation:

Ni se spune la întrebarea de mai sus să găsim suma depusă în bancă.

Ni se oferă următoarele valori

R = rata = 2% = 0.02

T = Timp = 1 an

Suma totală = A = 8160 lei

Trebuie să găsim principalul (suma depusă la bancă).

Formula de utilizat este dată ca:

P (Principal) = A / (1 + rt)

P = 8160 lei / 1 + 0.02 × 1

P = 8160 lei / 1.02

P = 8000 lei

Prin urmare, suma depusă în bancă este de 8000 de lei.

3 0

3 years ago