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bezimeni [28]
3 years ago
12

Ella and her children went into a bakery and where they sell cookies for $0.50 each and brownies for $2.25 each. Ella has $15 to

spend and must buy a minimum of 9 cookies and brownies altogether. If Ella decided to buy 5 cookies, determine the maximum number of brownies that she could buy. If there are no possible solutions, submit an empty answer.
Mathematics
2 answers:
Mice21 [21]3 years ago
6 0

Answer:

If each cookie is .50 and she wants to buy 5 cookies @ .50= $2.50

Than subtract $2,50 from $15.00 (which she has to spend you have $12.50

Then divide 2.25 into $12.5 and you get an answer of 5 brownies with a $1.30 extra to spend

Step-by-step explanation:

matrenka [14]3 years ago
5 0

Answer: 5 brownies

Step-by-step explanation:

cookies = 5 x $0.50 = $2.5

$15 - $2.5 = $12.5

$12.5 divided by $2.25 = $5.555

so at the most ella can buy 5 brownies if she buys 5 cookies

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Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c
GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(c) The value of P (X < 30) is 0.9502.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0

The value of E (X) is 10.

The parameter λ is:

\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10

(a)

Compute the value of P (X > 10) as follows:

P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353

Thus, the value of P (X > 20) is 0.1353.

(c)

Compute the value of P (X < 30) as follows:

P(X

Thus, the value of P (X < 30) is 0.9502.

(d)

It is given that, P (X < x) = 0.95.

Compute the value of <em>x</em> as follows:

P(X

Take natural log on both sides.

ln(e^{-0.10x})=ln(0.05)\\-0.10x=-2.996\\x=\frac{2.996}{0.10}\\ =29.96\\\approx30

Thus, the value of x is 30.

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Step-by-step explanation:

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Step-by-step explanation:

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LT1 10th grade level
Anton [14]

Answer:

The distance between the pirate and the treasure can be found from the following relationship between ΔEAF and ΔABC;

\overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC} = \overline{EF}/\overline{BC}

Step-by-step explanation:

From the question diagram, we have two triangles, ΔEAF and ΔABC;

The ratio of the lengths of the sides \overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC}

∠EAF and ∠BAC are vertical angles, therefore ∠EAF = ∠BAC

Therefore, ΔEAF and ΔABC are similar triangles by Side-Angle-Side, SAS, rule of similarity which states that two triangles that have ratios of a pair of their corresponding sides and the two sides also form equal angles within each triangle, then the two triangles are similar

Therefore, the ratio of the each pair of corresponding sides of the two triangles are equal

We have;

\overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC} = 50 ft. /(100 ft.) =  \overline{EF}/\overline{BC} = \overline{EF}/120 ft.

\overline{EF} = 120 ft. × 50 ft./(100 ft.) = 60 ft.

\overline{EF} = 60 ft.

The distance between the pirate and the treasure, \overline{EF} = 60 ft.

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