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2 years ago

What is 2x-y <4 graphed on a graph

Mathematics

1 answer:

photoshop1234 [79]2 years ago

7 0

Answer:

y>2x-4

y int = -4

over 2 right 1 (like staircase)

The line is dotted and shade to the left of graph

Step-by-step explanation:

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Anthony's Nuffin Shop recently sold 4 blackberry muffins and 6 other muffins. Considering

Mazyrski [523]

Answer:

Step-by-step explanation:

In total, Antony sold 10 muffins. We can do a ratio like this:

$\frac{4}{10} = \frac{x}{20}$

Because there are 4 blackberry muffins. We solve.

Because 10 is directly half of 20, we can just multiply 4 by 2 because if you multiply 10 by 2, you'd get 20. 4 times 2 is 8. That's the answer.

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3 years ago

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Answer:

28.25

Step-by-step explanation:

just do 226 divided by 8

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3 years ago

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3 years ago

luis is making two batches of muffins for a school picnic. one batch of muffins uses 1/3 cup of oats and 1/2 cup of flour. how m

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3 years ago

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At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$

$P( X \le 4 ) = 0.0054$

4 0

3 years ago