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3 years ago

ROBERT bought 2 packages of cookies for $6.00. How much would 3 packages of cookies cost?

Mathematics

1 answer:

babunello [35]3 years ago

5 0

Answer:

9.00$

Step-by-step explanation:

6 divided by 2 = 3 so each pakage cost 3$ so it would be 9.00$

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What is the simplified base for the function f(x) = 2rootindex 3 startroot 27 endroot superscript 2 x?

VMariaS [17]

Applying exponential properties, the simplified base for the function is given as follows:

$f(x) = 2(3)^{2x}$

<h3>What is the function?</h3>

The function is represented by the following definition:

$f(x) = 2\sqrt[3]{27^{2x}}$

As an exponent, the function is given by:

$f(x) = 2(27)^{\frac{2x}{3}}$

27 can be simplified as follows:

27 = 3 x 3 x 3 = 3³

Hence:

$f(x) = 2(3^3)^{\frac{2x}{3}}$

Then, the 3 at the exponent multiplies 2x and can be simplified with the 3 of the root, hence:

$f(x) = 2(3)^{\frac{3 \times 2x}{3}} = 2(3)^{2x}$

Hence the simplified base for the function is given as follows:

$f(x) = 2(3)^{2x}$

More can be learned about exponential properties at brainly.com/question/25263760

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2 years ago

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up

Artyom0805 [142]

Answer:

In the long run, ou expect to lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X $\sim$ geom(p = 1/2)

the probability function P can be computed as:

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

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid $\sum \limits ^{n}_{i=1} n^2 P(X=n)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2$ ∵ X $\sim$ geom(p = 1/2)

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =6$

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to lose $4 per game

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3 years ago

The graph of linear function g passes through the points (3, 7) and (9, 3).

Wewaii [24]

Answer:

Step-by-step explanation:

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