All categories

3 years ago

Solve 2|4w−1|=3|4w+2|.

Mathematics

1 answer:

Ivenika [448]3 years ago

5 0

Answer:

w = –2 or –1/5

Step-by-step explanation:

Solve the 2 possible equations:

2|4w–1| = 3|4w+2|

2|4w–1| = –3|4w+2|

You might be interested in

I need help the volume of a sphere is

Alenkinab [10]

$V=36\pi\ ft^3\\\\V=\dfrac{4}{3}\pi r^3\\\Downarrow\\\\\dfrac{4}{3}\pi r^3=36\pi\ \ \ |:\pi\\\\\dfrac{4}{3}r^3=36\ \ \ |\cdot3\\\\4r^3=108\ \ \ \ |:4\\\\r^3=27\to r=\sqrt[3]{27}\to r=3\ ft$

Answer: The radius r = 3 ft.

8 0

3 years ago

Given the functions k(x) = 5x − 8 and p(x) = x − 4, solve k[p(x)] and select the correct answer below.

Goryan [66]

Answer is B) k[p(x)] = 5x − 28

8 0

3 years ago

Read 2 more answers

The median of the data set is 81 what number is missing in data set 82, 87, 77, 70, 80, 88

Gekata [30.6K]

The number that missing is clearly 81

6 0

3 years ago

Read 2 more answers

Which choice could represent the volume of a cake pan

Sav [38]

Answer:

3150^3

Step-by-step explanation:

5 0

3 years ago

You are running a race. The probability that you win is 3/5. What is the probability that you lose at most 2 out of your 6 races

Veronika [31]

Answer:

The probability that you lose at most 2 out of your 6 races is 0.54432.

Step-by-step explanation:

We are given that you are running a race. The probability that you win is 3/5.

There are total of 6 races.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 6 races

r = number of success = at most 2 lost

p = probability of success which in our question is probability that

you lose a race = 1 - (3/5) = 2/5 or 0.4

Let X = Number of races lost

So, X ~ Binom(n = 6, p = 0.40)

Now, the probability that you lose at most 2 out of your 6 races is given by = P(X $\leq$ 2)

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

= $\binom{6}{0} \times 0.40^{0} \times (1-0.40)^{6-0} + \binom{6}{1} \times 0.40^{1} \times (1-0.40)^{6-1} + \binom{6}{2} \times 0.40^{2} \times (1-0.40)^{6-2}$

= $1 \times1 \times 0.60^{6} + 6 \times 0.40^{1} \times 0.60^{5} +15\times 0.40^{2} \times 0.60^{4}$

= 0.54432

7 0

3 years ago