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3 years ago

What is the solution to the system of equations y=-5x+3

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

8 0

Hola User______

Here is Your Answer...!!!!

_________________

thus the given equation is a linear equation in two variables ...

thus ploting a table we get as ...

y = -5x + 3

there fore

1) at y=0 , x = 3/5 .... 0 = -5x +3

= -5x = -3 ..thus X = 3/5

2) at x=0 , y= 3 ...

y = -5 (0) + 3 = 3

3) at X =1 , y = -2

4) at y=1 , x = 2/5

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Class of 50 students dressed up 38% have white shirts. How many are wearing white shirts?

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Step-by-step explanation:

☥ $\underline{ \underline{ \text{Given} }}:$

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A box contains 24 transistors,4 of which are defective. If 4 are sold at random,find the following probabilities. i. Exactly 2 a

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This is a binomial probability. For i, we will apply the Binomial probability formula

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Using the formula, we have

$\begin{gathered} P_x=^nC_x\left(p^x\right?\left(q^{n-x}\right) \\ Where\text{ } \\ P_x=binomial\text{ probability} \\ x=number\text{ of times for a specific outcome with n trials =2} \\ p=\text{ probability of success = }\frac{4}{24}=\frac{1}{6} \\ q=probability\text{ of failure =1-}\frac{1}{6}=\frac{5}{6} \\ ^nC_x=\text{ number of combinations = }^4C_2 \\ n=\text{ number of trials = 4} \end{gathered}$

Note that I made the probability of being defective as the probability of success = p

and probability of none defective as probability of failure = q

Exactly 2 are defective becomes the binomial probability

$\begin{gathered} P_x=^4C_2\times\lparen\frac{1}{6})^2\times\lparen\frac{5}{6})^{4-2} \\ P_x=6\times\frac{1}{36}\times\frac{25}{36} \\ P_x=\frac{25}{216} \\ =0.1157 \end{gathered}$

Hence the answer is 0.1157

(ii) None is defective becomes