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postnew [5]
3 years ago
14

4. The ages of the volunteers at a local food

Mathematics
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

The median is 43

Step-by-step explanation:

I dont know why your answers dont have any of them,cause with the numbers

3425, 24, 50, 18. 46. 43, 36.32

that makes your median is 43

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How to find the equivalent to 2/5​
Alex_Xolod [135]

Answer:

4/10, 6/15, 8/20, 10/25

Step-by-step explanation:

Those are the fractions equivalent to 2/5

8 0
3 years ago
Read 2 more answers
33 - (3.1 + 4.4) + 2.5<br> + 2.5<br> 33<br> Add.
OleMash [197]

<em>12</em>

<em>Step by Step, Following the rules of Pemdas</em>

<em>22 - (3.1 + 4.4) + 2.5</em>

<em>22 - 7.5 + 2.5</em>

<em>22 - 10</em>

<em>12</em>

<em />

<em>-Ɽ3₮Ɽ0 Ⱬ3Ɽ0 </em>

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3 0
3 years ago
6) Supplementary Exercise 5.51
tresset_1 [31]

Answer:

P(X \le 4) = 0.7373

P(x \le 15) = 0.0173

P(x > 20) = 0.4207

P(20\ge x \le 24)= 0.6129

P(x = 24) = 0.0236

P(x = 15) = 1.18\%

Step-by-step explanation:

Given

p = 80\% = 0.8

The question illustrates binomial distribution and will be solved using:

P(X = x) = ^nC_xp^x(1 - p)^{n-x}

Solving (a):

Given

n =5

Required

P(X\ge 4)

This is calculated using

P(X \le 4) = P(x = 4) +P(x=5)

This gives:

P(X \le 4) = ^5C_4 * (0.8)^4*(1 - 0.8)^{5-4} + ^5C_5*0.8^5*(1 - 0.8)^{5-5}

P(X \le 4) = 5 * (0.8)^4*(0.2)^1 + 1*0.8^5*(0.2)^0

P(X \le 4) = 0.4096 + 0.32768

P(X \le 4) = 0.73728

P(X \le 4) = 0.7373 --- approximated

Solving (b):

Given

n =25

i)

Required

P(X\le 15)

This is calculated as:

P(X\le 15) = 1 - P(x>15) --- Complement rule

P(x>15) = P(x=16) + P(x=17) + P(x =18) + P(x = 19) + P(x = 20) + P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)

P(x > 15) = {25}^C_{16} * p^{16}*(1-p)^{25-16} +{25}^C_{17} * p^{17}*(1-p)^{25-17} +{25}^C_{18} * p^{18}*(1-p)^{25-18} +{25}^C_{19} * p^{19}*(1-p)^{25-19} +{25}^C_{20} * p^{20}*(1-p)^{25-20} +{25}^C_{21} * p^{21}*(1-p)^{25-21} +{25}^C_{22} * p^{22}*(1-p)^{25-22} +{25}^C_{23} * p^{23}*(1-p)^{25-23} +{25}^C_{24} * p^{24}*(1-p)^{25-24} +{25}^C_{25} * p^{25}*(1-p)^{25-25}

P(x > 15) = 2042975 * 0.8^{16}*0.2^9 +1081575* 0.8^{17}*0.2^8 +480700 * 0.8^{18}*0.2^7 +177100 * 0.8^{19}*0.2^6 +53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0  

P(x > 15) = 0.98266813045

So:

P(X\le 15) = 1 - P(x>15)

P(x \le 15) = 1 - 0.98266813045

P(x \le 15) = 0.01733186955

P(x \le 15) = 0.0173

ii)

P(x>20)

This is calculated as:

P(x>20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)

P(x > 20) = 12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0

P(x > 20) = 0.42067430925

P(x > 20) = 0.4207

iii)

P(20\ge x \le 24)

This is calculated as:

P(20\ge x \le 24) = P(x = 20) + P(x = 21) + P(x = 22) + P(x =23) + P(x = 24)

P(20\ge x \le 24)= 53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1

P(20\ge x \le 24)= 0.61291151859

P(20\ge x \le 24)= 0.6129

iv)

P(x = 24)

This is calculated as:

P(x = 24) = 25* 0.8^{24}*0.2^1

P(x = 24) = 0.0236

Solving (c):

P(x = 15)

This is calculated as:

P(x = 15) = {25}^C_{15} * 0.8^{15} * 0.2^{10}

P(x = 15) = 3268760 * 0.8^{15} * 0.2^{10}

P(x = 15) = 0.01177694905

P(x = 15) = 0.0118

Express as percentage

P(x = 15) = 1.18\%

The calculated probability (1.18%) is way less than the advocate's claim.

Hence, we do not believe the claim.

5 0
3 years ago
The value of 7 in 67908 is blank times the value of 7 in 74702
Anastasy [175]
The value of 7 in 67098 is 7000 and the value of 7 in 74702 is 70000 or 700 so it is either 10 times or 1/10th of the the value.
4 0
3 years ago
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1 2/3 divided by 1 1/4
sashaice [31]

1 2/3 = 5/3

1 1/4 = 5/4       -----------> flip this one to make it multiplication

5/3*4/5 = 5*4 / 3*5 = 20/15 = 4/3 = 1 1/3

7 0
3 years ago
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