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kompoz [17]
3 years ago
10

Please answer this question now

Mathematics
2 answers:
iren [92.7K]3 years ago
6 0
JK is 3 inches.... hope this helps
Kaylis [27]3 years ago
5 0

Answer:

JK=  3 in

Step-by-step explanation:

There is a theorem that says that the angle of a tangent and the radius is of 90°

so we can use the pythagora's theorem

JK^2+JH^2=HK^2\\JK^2+(4)^2=(5)^2\\JK^2+16=25\\JK^2=9\\JK=3

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Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
  • n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

8 0
3 years ago
1. (k+7)² =289<br>2. (2s-1)² =225<br>3. (x-4)² =169<br><br>please help me answer these :)
professor190 [17]
(k+7)^2 =289\\&#10;|k+7|=17\\&#10;k+7=17\vee k+7=-17\\&#10;k=10 \vee k=-24

(2s-1)^2 =225\\&#10;|2s-1|=15\\&#10;2s-1=15\vee 2s-1=-15\\&#10;2s=16 \vee 2s=-14\\&#10;s=8 \vee s=-7

(x-4)^2 =169\\&#10;|x-4|=13\\&#10;x-4=13 \vee x-4=-13\\&#10;x=17 \vee x=-9
6 0
3 years ago
Read 2 more answers
How do I simplify the value of y= 200,000(1.06)15
Fudgin [204]

Answer:

For this, first you must multiply the exponent, which in this case is 1.06^15,

and this gives you 2.39655819.

You then multiply 2.39655819 by 200000.

y = 479311.638

4 0
3 years ago
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What is the acceleration due to gravity of a planet where a mass of 45.0 kg has a weight of
AnnZ [28]

Answer:

Normally, Acceleration due to gravity is

9.8m/s^2\\

or

10m/s^2

Step-by-step explanation:

From the web

These two laws lead to the most useful form of the formula for calculating acceleration due to gravity: g = G*M/R^2, where g is the acceleration due to gravity, G is the universal gravitational constant, M is mass, and R is distance

6 0
3 years ago
3. Given f(x)=-2(5)<br> Find f (4)
zzz [600]

Answer:

f

(

x

)

=

2

, means, whatever be the value of

x

, f(x)=2#.

∴

f

(

−

4

)

=

2

.

6 0
2 years ago
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