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3 years ago

In the formula p= mv what unit represents mass?

Mathematics

1 answer:

valentina_108 [34]3 years ago

7 0

Answer:

Step-by-step explanation:

Usually, the variable in a formula is the first letter of what it stands for.

Hope this helps!

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A rectangle has a length of 15.8 inches and a width of 11.9 inches. What is the area of the rectangle? Enter your answer as a de

pshichka [43]

Answer:

Hi, I belive your answer is 188.02

sorry if it is wrong

Here is how to get the answer which is 188.02; 11.9 x 15.8= 188.02in²

and here you go have a splendid day!

Step-by-step explanation:

6 0

3 years ago

Plz help me It’s due by tonight

natulia [17]

Answer:s

12 weeks

Step-by-step explanation:

if keith starts with $500 and wants to end with (at least) $200

he can withdrawn up to $500-$200=$300

if he withdraws $25 week the $300 will last $300 $25 week =12 weeks

5 0

3 years ago

Read 2 more answers

Solve the equation for x. Sqrt X-6+3 = 10 x=1 x=13 x = 43 x= 55

Jlenok [28]

Answer is 55
Two ways you can approach this first you can subtract the three from the 10 and get 7. Square that to get 49. Now you know that x-6 has to equal 49 and the only option greater than 49 is 55
Second you can just take the problem step by step.
Subtract three from both sides.
Sqrt x-6 =7
Square both sides
X-6=7^2
x-6=49
Add six to each side
X=55

8 0

3 years ago

Suppose that 7 in every 10 auto accidents involve a single vehicle. If 15 auto accidents are randomly selected, compute the prob

kobusy [5.1K]

Answer:

$\mathbf{P(X \le 4 ) \simeq 0.0006722}$

Step-by-step explanation:

From the information given:

p = x/n

p = 7/10

p = 0.7

sample size n = 15

Suppose X be the number of accidents involved by a single-vehicle.

Then;

$X \sim Binom (15,0.7)$

Thus, the required probability that at most 4 involve in a single-vehicle is

$P(X\le 4) \\ \\P(X \le 4) = P(X = 0) + P(X =1 ) + ... + P(X = 4)$

$P(X \le 4 ) = (^{15}_0) *0.7^0 *0.3^{15-0} + (^{15}_1) *0.7^1 *0.3^{15-1} + (^{15}_2) *0.7^2 *0.3^{15-2} + (^{15}_3) *0.7^3 *0.3^{15-3} + (^{15}_4) *0.7^4 *0.3^{15-4}$

$P(X \le 4 ) = (\dfrac{15!}{0!(15-0)!}) *0.7^0 *0.3^{15-0} + (\dfrac{15!}{1!(15-1)!}) *0.7^1 *0.3^{15-1} + (\dfrac{15!}{2!(15-2)!}) *0.7^2 *0.3^{15-2} + (\dfrac{15!}{3!(15-3)!}) *0.7^3 *0.3^{15-3} + (\dfrac{15!}{4!(15-4)!}) *0.7^4 *0.3^{15-4}$

$P(X \le 4 ) =1.4348907 \times 10^{-8} +5.02211745 \times 10^{-7} + 8.20279183 \times 10^{-6} + 8.29393397 \times 10^{-5} + 5.80575378 \times 10^{-4}$

$P(X \le 4 ) =6.7223407 \times 10^{-4}$

$\mathbf{P(X \le 4 ) \simeq 0.0006722}$

3 0

3 years ago

Write an equation of the line that passes through the points (−3,−4) and (0,2). y=

AysviL [449]

Answer:

y=2x+2

Step-by-step explanation:

3 0

3 years ago