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2 years ago

What formula do I use for this

Mathematics

1 answer:

hichkok12 [17]2 years ago

5 0

Answer:

Distance=520cm

Step-by-step explanation:

The solution is in the image

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The function g(t) = - 16t2 + vt + h represents the height of an object, g, in feet, above the ground in relation to the time, t,

timurjin [86]

Answer:

$g(t) = -16t^2+80t$ .

Step-by-step explanation:

We are given that $v=80$ and $h=0$ (because that's the ground), so substituting that into $g(t) = -16t^2+vt+h$ gives $g(t) = -16t^2+80t$ .

5 0

2 years ago

Read 2 more answers

Tan9 - tan27 - tan63 + tan 81 = ?

ludmilkaskok [199]

Tan9−tan27−tan63−tan81
tan9+tan81−tan27−tan63
sin9/cos9+sin81/cos81−sin27/cos27−sin63/cos63
sin90/cos81cos9−sin90/cos63cos27
1/sin9cos9−1/sin27cos27
2/sin18−2/sin54
(2)sin54−sin18/sin18sin54
4cos36sin18/sin18cos36=4

8 0

3 years ago

Where can the medians of a triangle intersect

gizmo_the_mogwai [7]

In side the triangle. Each median is split in the ratio 2:1 by the intersection.

6 0

3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

Allana 3/5 used yard of fabric to make a scarf. Can she make 2 of these scarves with 1 7/10 yards of fabric, and why?

Andreas93 [3]

One scarf takes 3/5 of a yard. Two of these takes 6/5 of a yard. If we put the 1 7/10 into an improper fraction we would get 17/10. So in order to compare the two fractions, the 6/5 and the 17/10, they have to have the same denominator. 6/5 can be rewritten as 12/10. 12/10 is less that the 17/10 you have, so yes you can make two scarves with that amount of fabric.

5 0

3 years ago