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3 years ago

If elephant grass grows 2 1/2 inches a day, how many inches will it grow in 9 days?

Mathematics

2 answers:

sergeinik [125]3 years ago

8 0

Just multiply 2 1/2 times 9, and there will be ur answer

Vikki [24]3 years ago

6 0

Is that 21/2 or 1/2 huh

elephant grass grows to 9/2 inches in 9 days

elephant grass grows to 4.5 inches in 9 days

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Which equation represents the line that is perpendicular to the graph of 4x + 3y = 9 and passes through (-2, 3)? RE A 3x - 4y =

Keith_Richards [23]

Answer:

15279

Step-by-step explanation:

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3 years ago

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Anastaziya [24]

-4, +1

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4 years ago

Para una función de teatro para niños, el boleto para adultos cuesta $70 y el boleto para niños $35. Anoche entraron en total 80

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3 years ago

Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual

Stells [14]

Answer:

(a) P( A ∩ B )=0.5 is not possible.

(b) 0.7

(d) 0.3

(e) 0.4

Step-by-step explanation:

Given information: The alphabet A and B represents the following events

A : Individual has a Visa credit card.

B: Individual has a MasterCard.

P(A)= 0.6 and P(B)=0.4.

(a)

We need to check whether P( A ∩ B ) can be 0.5 or not.

$A\cap B\subset A$ and $A\cap B\subset B$

$P(A\cap B)\leq P(A)$ and $P(A\cap B)\leq P(B)$

$P(A\cap B)\leq 0.6$ and $P(A\cap B)\leq 0.4$

From these two inequalities we conclude that

$P(A\cap B)\leq 0.4$

Therefore, P( A ∩ B )=0.5 is not possible.

(b)

Let $P(A\cap B)=0.3$

We need to find the probability that student has one of these two types of cards.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Substitute the given values.

$P(A\cup B)=0.6+0.4-0.3=0.7$

Therefore the probability that student has one of these two types of cards is 0.7.

(c)

We need to find the probability that the selected student has neither type of card.

$P(A'\cup B')=1-P(A\cup B)$

$P(A'\cup B')=1-0.7=0.3$

Therefore the probability that the selected student has neither type of card is 0.3.

(d)

The event that the select student has a visa card, but not a mastercard is defined as

$A-B$

It can also written as

$A\cap B'$

The probability of this event is

$P(A\cap B')=P(A)-P(A\cap B)$

$P(A\cap B')=0.6-0.3=0.3$

Therefore the probability that the select student has a visa card, but not a mastercard is 0.3.

(e)

We need to find the probability that the selected student has exactly one of the two types of cards.

$P(A\cap B')+P(A\cap B')=P(A\cup B)-P(A\cap B)$

$P(A\cap B')+P(A\cap B')=0.7-0.3$

$P(A\cap B')+P(A\cap B')=0.4$

Therefore the probability that the selected student has exactly one of the two types of cards is 0.4.

8 0

3 years ago

Using the Descartes Rule of Signs, describe the real zeroes of the function.

mrs_skeptik [129]

as you already know, Descartes rule of signs check the times the sign changes or f(x) and for f(-x)

$\bf \stackrel{\textit{positive roots}}{f(x)=}\underset{change}{2x^5-}x^4\underset{change}{-2x^3+}4x^2\underset{change}{+x-2} \\\\\\ \stackrel{\textit{negative roots}}{f(~-x~)=}-2x^5\underset{change}{-x^4+}2x^3\underset{change}{+4x^2-}x-2$

by the fundamental theorem of algebra, the polynomial has a degree of 5, so it has to have at most 5 zeros/solutions/roots.

f(x) has 3 sign changes, notice, that means, it has either 3, or (3-2), 1 positive zeros.

for f(-x), recall that x³, will be (-x)³ = (-x)(-x)(-x) = -x³, so in short, if the exponent is ODD, the sign changes for that term, if it's EVEN, it doesn't change.

so for f(-x), we have 2 sign changes, meaning, it has either 2 or (2-2), 0 negative roots.

the slack is picked up by the complex roots.

3 positive, 2 negative, 0 complex

1 positive, 2 negative, 2 complex *recall complex always come in pairs*

3 positive, 0 negative, 2 complex

1 positive, 0 negative, 4 complex.

5 0

3 years ago