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natali 33 [55]
3 years ago
15

A triangle is made up of two sides that are 8 cm long and 12 cm long which of the following could be the length of the third sid

e? (select all possible answers)
A. 8cm
B. 2cm
C. 28 cm
D. 15 cm
E. 19 cm
Mathematics
1 answer:
Levart [38]3 years ago
3 0
I think a b and c , not positive
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2 2/5

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Sin(x-pi/6)+cos(x-pi/3)=(√3+1)÷4​
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4 years ago
A population of deer inside a park has a carrying capacity of 200 and a growth rate of 3%. If the initial population is 80 deer,
otez555 [7]

Answer:

The population of deer at any given time = 200(e^0.03t)  ÷ (1.5 + (e^0.03t))

Step-by-step explanation:

This is an example of logistic equation on population growth

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P(t) =?

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P(t) = (80 (200)(e^0.03t)) ÷ (200 - 80 + 80(e^0.03t))

     = (16000(e^0.03t))  ÷ (120 + 80(e^0.03t))

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8 0
3 years ago
Please help I will give out brainliest
kolbaska11 [484]

Hey There!! ~

The answer to this is: the upper bound for the length is 21.5cm. Lower and Upper Bounds

The lower bound is the smallest value that will round up to the approximate value.

The upper bound is the smallest value that will round up to the next approximate value.

Ex:- a mass of 70 kg, rounded to the nearest 10 kg, The upper bound is 75 kg, because 75 kg is the smallest mass that would round up to 80kg.

Here , A length is measured as 21cm correct to 2 significant figures. We need to find what is the upper bound for the length . let's find out:

As discussed above , upper bound for any number will be the smallest value in decimals which will round up to next integer value . So , for 21 :

⇒  21.5cm.

21.5 cm on rounding off will give 22 cm . So , the upper bound for the length is 21.5cm.

Hope It Helped!~

ItsNobody~

7 0
3 years ago
In a survey of women in a certain country the mean height was 62.9 inches with a standard deviation of 2.81 inches answer the fo
Natasha2012 [34]

The question is incomplete. The complete question is :

In a survey of women in a certain country ( ages 20-29), the mean height was 62.9 inches with a standard deviation of 2.81 inches.  Answer the following questions about the specified normal distribution.  (a) What height represents the 99th percentile?  (b) What height represents the first quartile?  (Round to two decimal places as needed)

Solution :

Let the random variable X represents the height of women in a country.

Given :

X is normal with mean, μ = 62.9 inches and the standard deviation, σ = 2.81 inches

Let,

$Z=\frac{X - 62.9}{2.81}$ , then Z is a standard normal

a). Let the 99th percentile is = a

The point a is such that,

$P(X

$P \left( Z < \frac{a-62.9}{2.81} \right) = 0.99$

From standard table, we get : P( Z < 2.3263) =0.99

∴ $\frac{(a-62.9)}{281} = 2.3263$

  $a= (2.3263 \times 2.81 ) +62.9$

     = 6.536903 + 62.9

     = 69.436903

     = 69.5 (rounding off)

Therefore, the height represents the 99th percentile = 69.5 inches.

b). Let b = height represents the first quartile.

It is given by :

P( X < b) =0.25

$P \left( Z < \frac{(b-62.9)}{2.81} \right) = 0.25$

From the standard normal table,

P( Z < -0.6745) =0.99

∴ $\frac{(b-62.9)}{2.81}= 0.6745$

$b=(0.6745 \times 2.81) +62.9$

  = 1.895345 + 62.9

   = 64.795345

   = 64.8 (rounding off)

Therefore, the height represents the 1st quartile is 64.8 inches.

8 0
3 years ago
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