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Alex777 [14]
3 years ago
5

Find the area of the figure to the nearest tenth.

Mathematics
2 answers:
kicyunya [14]3 years ago
8 0
Area is length x width
Andreas93 [3]3 years ago
3 0
To find the area of a square or rectangle, it is length time width. 
To find the area of a triangle it is .5base times height.
To find the area of a parallelogram, it is base times height
To find the area of a circle, it is pi times r squared. 
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Help me ASAP please
Montano1993 [528]

Answer:  i feel that the coorect answer is b

6 0
3 years ago
Read 2 more answers
Question 17
Rufina [12.5K]

Answer:

698 fishes

Step-by-step explanation:

Generally, we can represent an exponential growth function as;

y = a•(1 + r)^t

originally, there were 3 fishes

The original value in this case means a = 3

After 6 weeks, there were 31

31 in this case is y

r is the increase percentage or rate

t is the time

So, we have it that;

31 = 3•(1 + r)^6

31/3 = (1 + r)^6

10.33 = (1 + r)^6

ln 10.33 = 6 ln (1 + r)

ln 10.33/6 = ln (1 + r)

e^0.3892 = (1 + r)

1 + r = 1.476

r = 1.476-1

r = 0.476 or 47.6%

So the growth percentage or rate is 47.6%

For 14 weeks, we simply have the value of t as 14;

So ;

y = 3•(1 + 0.476)^14

y = 3(1.476)^14

y = 698 fishes

8 0
3 years ago
PLS HELP WITH THIS QUESTION GIVING BRAINLIST
Aleks04 [339]

Answer: 1: 5(3x+4)

2: 6(x2+4x−3)

3: 5x(4x2−3x+2)

4: 4n2(4n2-3)

Step-by-step explanation:

8 0
3 years ago
T had 53,790 miles onit. They had the car for 8.25 years. On average, how many miles did they drive per year?
liq [111]

Answer: ‭6,520‬ miles per year

Step-by-step explanation:

They had driven the car for 53,790 miles in 8.25 years.

The average number of number of miles driven per year is:

= 53,790 / 8.25 years

= ‭6,520‬ miles per year

<em>They drove an average of ‭6,520‬ miles per year.</em>

7 0
2 years ago
the mgf of a random variable x is e^3(e^t-1). Find P[mean - standard deviation squared &lt; X &lt; 1/2( mean + standard deviatio
postnew [5]
The given MGF is that for a random variable following a Poisson distribution with parameter \lambda=3.

This means \mathbb E(X)=\mathbb V(X)=\lambda, and X has PMF

f_X(x)=\begin{cases}\dfrac{3^xe^{-3}}{x!}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}

So, the desired probability is

\mathbb P\left(\lambda-\lambda^2

This is equivalent to

\displaystyle\sum_{x=0}^2\mathbb P(X=x)=\sum_{x=0}^2\frac{3^x}{x!e^3}=\frac{17}{2e^3}\approx0.4232
8 0
3 years ago
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