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Tanzania [10]
3 years ago
6

The graph below shows the winter precipitation in Arizona from 1995 to 2002. It also shows the survival rates of Arizona prongho

rn fawns for the same period.
Which two years had the most precipitation?

• Which two years had the best survival rates for fawns?
• Using your knowledge of biotic and abiotic factors and information from the graph, give two reasons why precipitation and fawn population are linked.
• Predict what would happen to the fawn population if the precipitation rate continues to go downward in 2003.

Mathematics
2 answers:
Anna [14]3 years ago
4 0
1998 and 2001 had the most precipitation


1998 and 2001 had the best survival rates for fawns


1) plants require water to grow, fawns feed of plants
2) they need water to drink


If it keeps going down we will no longer have fawns in the future
Nana76 [90]3 years ago
3 0

Answer:

* 1998 and 2001 had the most precipitation.

* The best years for survival were 1998 and 2001.

* One reason would be that precipitation rates are connected to plants as well. Fawns are herbivores, meaning they eat plants, so if the plants are well nourished by rain and growing, fawns will have plenty to eat. Sorry that I don’t know the second reason the two factors are connected.

* Fawns would most likely go extinct if the precipitation rate continues to go down.

Step-by-step explanation:

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Find the following bases.
Oduvanchick [21]

Answer:

a.. 8% of ? = 20   <u> 240</u>

b.75% of ? = 30    <u>40</u>

c. 20% of ? = 45   <u>225</u>

d. 150% of ? = 36  <u>24</u>

The steps of  answer is in attached file

Hope it helps..

Have a great day : )

6 0
3 years ago
Read 2 more answers
There are five seniors in a class. For each situation, write how the binomial formula is used to calculate the probability.
ololo11 [35]
A) The answer is 5.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
<span>I choose one senior: r = 1
</span>
nCr = n! / (r! (n - r)!)
5C1 = 5! / (1! (5 - 1)!)
       = (5 * 4 * 3 * 2 * 1) / (1 * 4!)
       = 120 / (4 * 3 * 2 * 1)
       = 120 / 24
       = 5

b) The answer is 10.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose two seniors: r = 2

nCr = n! / (r! (n - r)!)
5C2 = 5! / (2! (5 - 2)!)
       = (5 * 4 * 3 * 2 * 1) / ((2 * 1) * 3!)
       = 120 / (2 * (3 * 2 * 1))
       = 120 / (2 * 6)
       = 120 / 12
       = 10


c) The answer is 10.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose three seniors: r = 3

nCr = n! / (r! (n - r)!)
5C3 = 5! / (3! (5 - 3)!)
       = (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * 2!)
       = 120 / (6 * (2 * 1))
       = 120 / (6 * 2)
       = 120 / 12
       = 10


d) The answer is 5.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose four seniors: r = 4

nCr = n! / (r! (n - r)!)
5C4 = 5! / (4! (5 - 4)!)
       = (5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * 1!)
       = 120 / (24 * 1)
       = 120 / 24
       = 5


e) The answer is 1.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose five seniors: r = 5

nCr = n! / (r! (n - r)!)
5C5 = 5! / (5! (5 - 5)!)
       = (5 * 4 * 3 * 2 * 1) / ((5 * 4 * 3 * 2 * 1) * 1!)
       = 120 / (120 * 1)
       = 120 / 120
       = 1
8 0
3 years ago
For each of the finite geometric series given below, indicate the number of terms in the sum and find the sum. For the value of
kotykmax [81]

Answer:

Number of term N = 9

Value of Sum = 0.186

Step-by-step explanation:

From the given information:

Number of term N = 3 (0.5)^{5} + 3 (0.5)^{6} + 3 (0.5)^{7} + \cdots + 3 (0.5)^{13}

Number of term N = 3 (0.5)^{5} + 3 (0.5)^{6} + 3 (0.5)^{7} +3 (0.5)^{8}+3 (0.5)^{9} +3 (0.5)^{10} +3 (0.5)^{11}+3 (0.5)^{12}+ 3 (0.5)^{13}

Number of term N = 9

The Value of the sum can be determined by using the expression for geometric series:

\sum \limits ^n_{k=m}ar^k =\dfrac{a(r^m-r^{n+1})}{1-r}

here;

m = 5

n = 9

r = 0.5

Then:

\sum \limits ^n_{k=m}ar^k =\dfrac{3(0.5^5-0.5^{9+1})}{1-0.5}

\sum \limits ^n_{k=m}ar^k =\dfrac{3(0.03125-0.5^{10})}{0.5}

\sum \limits ^n_{k=m}ar^k =\dfrac{(0.09375-9.765625*10^{-4})}{0.5}

\sum \limits ^n_{k=m}ar^k =0.186

6 0
3 years ago
YOU DONT NEED TO DO ANYTHING JUST LET ME KNOW IF IM RIGHT OR NOT AND GIVE ME A REASON WHY IM CORRECT OR INCORRECT FOR 10 POINTS!
jeyben [28]

incoorect Step-by-step explanation:

Supplementary Angles  two angles that sum up to 180°

the other two sides are that and the other side is 70

7 0
3 years ago
What is the overlap of Data Set 1 and Data Set 2?
erastovalidia [21]

Answer: Answer:

There is moderate level of overlap between the two data sets.  

Step-by-step explanation:

Overlap of data sets.

Overlap measures the degree of duplication that exists within data sets.

It is an indicator of the degree to which data are identical.

We are given two data sets in the question.

We have to find the amount of overlap in data set 1 and in data set 2.

There are 8 data points in data set 1 as shown in the image.

There are 8 data points in the data set 2 as shown in the image.

Out of the 8 data points for both data set 1 and data set 2, 5 data points overlap each other on 6, 7, 8 and 9.

Thus, we could say there is moderate level of overlap between the two data sets.

3 0
2 years ago
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