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2 years ago

Can anyone please help me

Mathematics

1 answer:

Dima020 [189]2 years ago

5 0

don't

stay in school kids please don't

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Create a factorable polynomial with a GCF of 2y. Rewrite that polynomial in two other equivalent forms. Explain how each form wa

Nezavi [6.7K]

2y^2 = x^2

y = x/sqrt(2)

how?

y = (x^2)/(2y)

y(2y) = (x^2)(2y)/(2y)

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2 years ago

Please help. im pretty stumped

sukhopar [10]

Answer:

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Step-by-step explanation:

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In quadrilateral ABCD, diagonals AC And BD bisect one another

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Segment AP is congruent to segment CP. Segment BP is congruent to segment AP Sides AB and BC are congruent. Triangles BCP and CDP are congruent.

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3 years ago

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D option that is 0 will be answer

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2 years ago

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A street magician asks a passerby to draw one card from a standard deck, remember it, and then replace it. The magician then shu

Romashka-Z-Leto [24]

Answer:

P( At least one of the two cards was a heart)=7/16

Step-by-step explanation:

The solution to this question can be found in two ways.

First way: Since total number of heart in a deck are four,

Probability that neither of two cards had a heart is that at both times the card drawn was not a heart, since it is an independent event, thus probability both were not heart= $\frac{3}{4}$ × $\frac{3}{4}$

= $\frac{9}{16}$

Thus, Probability that at least one of the two cards was a heart= 1- P( Neither of two cards had a heart= 1- $\frac{9}{16}$

= $\frac{7}{16}$

Second way:

Probability that first card drawn is a heart and the second one is not a heart= $\frac{1}{4}$ × $\frac{3}{4}$

= $\frac{3}{16}$

Probability that the first card drawn is not a heart and the second one is a heart= $\frac{3}{4}$ × $\frac{1}{4}$

= $\frac{3}{16}$

Probability that both cards drawn are hearts= $\frac{1}{4}$ × $\frac{1}{4}$

= $\frac{1}{16}$

Adding all these probabilities, we have the probability that at least one of the card drawn was heart= $\frac{7}{16}$

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3 years ago