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Oksana_A [137]
3 years ago
12

OF FRACTIONS

Mathematics
1 answer:
kodGreya [7K]3 years ago
7 0

Answer:

<em>qsyyyyyy                                                                                                          ;-;                   </em>

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The items Matt ate at dinner cost m dollars. He added 6.5% to the cost of dinner for sales tax. Matt paid $25 using a gift card
Artyom0805 [142]

Answer:

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Step-by-step explanation:

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3 years ago
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Find the x- and y-intercept of the line with equation -3x+9y=18
ratelena [41]
Plug in “0” for y, rearrange and solve.
Remember when you want to find the intercept of a variable, you plug in 0 for the opposite variable. Same goes for finding the x intercept.

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Answers:
X int= -6 or (-6,0)
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3 years ago
5.346 - 2.289 = ? <br><br> SHOW WORK <br><br> WILL GIVE BRAINLIEST PLEASE
Vlad1618 [11]

Answer:

7.635

Step-by-step explanation:

1 1  

 |5| . |3| |4| |6|

+  |2| . |2| |8| |9|

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3 years ago
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4 years ago
Consider a binomial experiment with n = 20 and p = .70.
jolli1 [7]

Complete question:

Consider a binomial experiment with n = 20 and p = .70.

A. Compute f(12).

B. Compute f(16).

C. Compute P(x≥ 16).

D. Compute P(x≤15).

E. Compute E(x).

F. Compute Var(x).

Answer:

a) 0.1144

b) 0.1304

c) 0.2375

d) 0.7625

e) 14

f) 4.2

Step-by-step explanation:

Given:

n = 20

p = 0.70

q = 1 - p ==>  1 - 0.70 = 0.30

a) Use the formula:

P(x) = CC\left(\begin{array}{ccc}n\\x\end{array}\right) p^x q^(^n^-^x^)

Thus,

P(12) = C\left(\begin{array}{ccc}20\\12\end{array}\right) (0.7^1^2) (0.3^(^2^0^-^1^2^) )

= 125970*0.0138*0.00006

= 0.1144

b) P(16) = C\left(\begin{array}{ccc}20\\16\end{array}\right) 0.7^1^6 (0.3^(^2^0^-^1^6^))

= 4845 * 0.0033 * 0.0081

= 0.1304

c) Compute P(x≥16):

P(x ≥ 16) = P(16) + P(17) + P(18) + P(19) + P(20)

= C\left(\begin{array}{ccc}20\\16\end{array}\right) 0.7^1^6 (0.3^(^2^0^-^1^6^)) + C\left(\begin{array}{ccc}20\\17\end{array}\right) 0.7^1^7 (0.3^(^2^0^-^1^7^) ) + C\left(\begin{array}{ccc}20\\18\end{array}\right) 0.7^1^8 (0.3^(^2^0^-^1^8^)) + C\left(\begin{array}{ccc}20\\19\end{array}\right) 0.7^1^9 (0.3^(^2^0^-^1^9^)) + C\left(\begin{array}{ccc}20\\20\end{array}\right) 0.7^2^0 (0.3^(^2^0^-^2^0^))

= 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.2375

d) P(x ≤ 15):

= 1 - P(x ≥ 16)

= 1 - 0.2375

= 0.7625

e) E(x): use the formula, n * p.

= n*p

= 20 * 0.7

= 14

f) Var(x)

Use the formula: npq

npq = 20 * 0.7 * 0.3

= 4.2

σ

5 0
4 years ago
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