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3 years ago

In 60 seconds, a turtle’s heart beats 20 times. To find the amount of time between heartbeats, what should you do?

Mathematics

1 answer:

Mars2501 [29]3 years ago

3 0

Answer:

1 heart beat every 3 seconds

Step-by-step explanation:

60 seconds

20 heartbeats

60/20=3

So 1 every 3 second

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Please Help!, Brainliest to whoever helps

Tanya [424]

B for sure!!!!!!!!!!!!!!!!!!!

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

Need help asap! Show work if you can, it’s completely ok if you don’t tho

Savatey [412]

Its A) -67
plug the m and n values into the function and solve using pemdas.
5(-7)-2(-7+3)^2
-35-2(-4)^2
-35-2(16)
-35-32
-67

8 0

3 years ago

I need to give an equation for an absolute value function that has a minimum. How do I determine that it has a minimum?.

tatyana61 [14]

To determine the minimum of an equation, we derive the equation using differential calculus twice (or simply take the second derivative of the function). If the second derivative is greater than 0, then it is minimum; else, if it is less than 1, the function contains the maximum. If the second derivative is zero, then the inflection point is identified.

5 0

3 years ago

If (x + 2) is a factor of x3 − 6x2 + kx + 10, k =

ch4aika [34]

Answer:

The value of k is -11

Step-by-step explanation:

If (x+2) is a factor of x3 − 6x2 + kx + 10:

Then,

f(x)=x3 − 6x2 + kx + 10

f(-2)=0

f(-2)=(-2)³-6(-2)²+k(-2)+10=0

f(-2)= -8-6(4)-2k+10=0

f(-2)= -8-24-2k+10=0

Solve the like terms:

f(-2)=-2k-22=0

f(-2)=-2k=0+22

-2k=22

k=22/-2

k=-11

Hence the value of k is -11....

5 0

3 years ago