Siobhan performs the elementary row operation represented by 2R3-R2 on matrix A.

The formula to determine energy is uppercase E = one-half m v squared. What is the formula solved for v?

alexandr402 [8]

Answer:

Step-by-step explanation:

$E=\frac{1}{2}mv^2$

All the variables on the right are being multiplied together then the whole mess is being divided by 2. Let's get rid of the 2 first. The undoing of division is multiplication, so we will begin by multiplying both sides by 2 to get

$2E=mv^2$

Next we will move the m. The undoing of multiplication is division. So we divide both sides by m to get

$\frac{2E}{m}=v^2$

The undoing of a square is to take the square root, so we will do that to both sides giving us, finally

$\sqrt{\frac{2E}{m} }=v$

6 0

3 years ago

Read 2 more answers

I need help please...

Ostrovityanka [42]

Answer:

B

Step-by-step explanation:

I think it's hitting a golf ball in the air, as it is a transformation, but the general shape of it is not affected.

Opening a locker, you are still affecting the general shape.

6 0

3 years ago

Please answer this question with any of the 4 choices i have been struggling with this

kotegsom [21]

Answer:

Choice A. $\frac{2}{3}$

Step-by-step explanation:

Turn the mixed numbers into improper fractions:

$\frac{14}{9} /\frac{7}{3}$

Flip the second term and multiply:

$\frac{14}{9} /\frac{3}{7}$
$\frac{42}{63}$

Simplify:

$\frac{2}{3}$

7 0

2 years ago

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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$