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meriva
4 years ago
9

On the day she was born, a baby leopard had 9 spots. Each day 4 more appeared. After 5 days, how many spots did the baby leopard

have?
Mathematics
2 answers:
kherson [118]4 years ago
7 0
The baby leopard would have no more spots because 4×5=20 and if the leopard had 9 spots at the start and over the course of five days it lost 20 spots, it would be -11 spots left.
Zolol [24]4 years ago
7 0
-11 spots would be the answer
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Novosadov [1.4K]
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Gretchen has a set of blocks of heights 1, 2, and 4-centimeters. Imagine stacking the blocks one on top of the other to make a t
notsponge [240]

Answer:

t_{n}=t_{n-1}+t_{n-2} +t_{n-4}

Step-by-step explanation:

t_{n}=multiple ways to climb a tower

When n = 1,

tower= 1 cm  

t_{1}= 1

When n = 2,

tower =2 cm   

t_{2}= 2

When n = 3,

tower = 3 cm

it can be build if we use three 1 cm blocks

t_{3} = 3

When n = 4,

tower= 4 cm

it can be build if we use four 1 cm blocks

t_{4} = 6

When n > 5

tower height > 4 cm

so we can use 1 cm, 2 cm and 4 cm blocks

so in that case if our last move is 1 cm block then t_{n-1} will be

n —1 cm

if our last move is 2 cm block then t_{n-2} will be

n —2 cm

if our last move is 4 cm block then t_{n-4} will be

n —4 cm

 

t_{n}=t_{n-1}+t_{n-2} +t_{n-4}

4 0
4 years ago
Read 2 more answers
Please help me <br> Find f(-4)
Katyanochek1 [597]
As x < 3, you would use the equation x^{2} -5
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8 0
3 years ago
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The median of a probability distribution can be defined as the number m such that Upper P (Upper X less than or equals m )equals
Alexxandr [17]

Answer:

tex]M=\beta ln(2)[/tex]

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate).

Solution to the problem

For this case we can use the following Theorem:

"If X is a continuos random variable of the exponential distribution with parameter \beta for some \beta \in R >0"

Then the median of X is \beta ln (2)

Proof

Let M the median for the random variable X.

From the definition for the exponential distribution we know the denisty function of X is given by:

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Since we need the median we can put this equation:

P(X

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\frac{1}{\beta} \int_0^M e^{- \frac{x}{\beta}}dx =\frac{1}{\beta} [-\beta e^{-\frac{x}{\beta}}] \Big|_0^M

And that's equal to:

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And if we solve for M we got:

1-e^{- \frac{M}{\beta}} = \frac{1}{2}

e^{- \frac{M}{\beta}}=\frac{1}{2}

If we apply natural log on both sides we got:

-\frac{M}{\beta}=ln(1/2)

And then M=\beta ln(2)

5 0
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