I forgot someone help me plz

Mathematics

2 answers:

baherus [9]3 years ago

8 0

1st Method:

(3² -2²) / (3-2)² = (9-4) / (1²) = 5

2nd Method:

(3² -2²) / (3-2)² = (3-2)(3+2) / (3-2)², after simplification by (3-2), the remaining numerator = (3+2) n= 5

o-na [289]3 years ago

5 0

Write the conjugates of them then FOIL both the top and bottom, then factor

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Solve for w. <br> 1/7 = -3/2w -4/5 Simplify your answer as much as possible.

Ann [662]

1/7 = -3/2w -4/5
add 4/5 to both sides
1/7 + 4/5= -3/2w
change to common denominator; 5 & 7 go into 35
(1*5)/35 + (4*7)/34= -3/2w
5/35 + 28/35= -3/2w
33/35= -3/2w
divide by -3/2 on both sides; multiply by the reciprocal of -3/2
33/35 * -2/3= w
-66/105= w
simplify by 3
-22/35= w

CHECK:
1/7 = -3/2w - 4/5
1/7= -3/2(-22/35) - 4/5
1/7= 66/70 - 4/5
1/7= 66/70 - 56/70
1/7= 10/70
1/7= 1/7

ANSWER: w= -22/35

Hope this helps! :)

5 0

3 years ago

What is the probability against a spinners landing o a green section if the spinner had 15 sections & 3 of them are green an

olganol [36]

The percentage would be 67%

5 0

3 years ago

An automobile manufacturer obtains the microprocessors used to regulate fuel consumption in its automobiles from three microelec

Anastaziya [24]

Answer:

The probability that a randomly selected automobile manufactured by the company will have a defective microprocessor is 6.2 %.

Step-by-step explanation:

Given:

The probability of getting a microprocessor from firm A is, $P(A)=45\%=0.45$

The probability of getting a microprocessor from firm B is, $P(B)=30\%=0.30$

The probability of getting a microprocessor from firm C is, $P(C)=25\%=0.25$

Now, let event D be having a defective microprocessor at random.

So, as per the question,

Probability of producing a defective microprocessor from firm A is, $P(D/A)=6\%=0.06$

Probability of producing a defective microprocessor from firm B is, $P(D/B)=7\%=0.07$

Probability of producing a defective microprocessor from firm C is, $P(D/C)=5.5\%=0.055$

Now, probability of having a defective microprocessor when selected at random is given as:

$P(D)=P(A)\cdot P(D/A)+P(B)\cdot P(D/B)+P(C)\cdot P(D/C)\\P(D)=(0.45\times 0.06)+(0.30\times 0.07)+(0.25\times 0.055)\\P(D)=0.027+0.021+0.01375\\P(D)=0.06175\ or\ P(D)=0.06175\times 100\approx 6.2\%$