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1 year ago

Last weekend, Lena worked 7.5 hours on Friday, 9.75 hours on Saturday, and 6.25 hours on Sunday.

Mathematics

1 answer:

Gemiola [76]1 year ago

3 0

Answer:202.1

Step-by-step explanation:

Add all the days she worked up (7.5+9.75+6.25) =23.5

Now simply just multiply it with 8.60 and you get your answer

Hope this helps :)

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In a West Texas school district the school year began on August 1 and lasted until May 31. On August 1 a Soft Drink company inst

WINSTONCH [101]

Answer:

$95.78

Step-by-step explanation:

f(t) = 300t / (2t² + 8)

t = 0 corresponds to the beginning of August. t = 1 corresponds to the end of August. t = 2 corresponds to the end of September. So on and so forth. So the second semester is from t = 5 to t = 10.

$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt

$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt

$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt

$T₂ = 75 ln(t² + 4) |₅¹⁰

$T₂ = 75 ln(104) − 75 ln(29)

$T₂ ≈ 95.78

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2 years ago

It costs dylan $x$ dollars to buy $3$ tickets to the zoo. he discovers that if he were to buy $5$ tickets, he would receive a to

jonny [76]

So, if Dylan has x dollars and he bought 3 tickets with them, the tickets were priced at k dollars per ticket. If he bought 5 tickets with the x dollars and saved 12 total dollars, it would be the same as buying the tickets with x-12 dollars, so we have:

$\frac{x}{3}=k\\ \frac{x-12}{5}=q$

So, with this we have:

$x=3k\\ x=5q+12$

If we're looking for a number that satisfies these constraints, we can work with modular arithmetic. We have:

$x\equiv 0 \pmod{3}\\ x\equiv 12 \pmod{5}$

So, we can use the chinese remainder theorem here. So, we clearly have x=3k, which means:

$3k\equiv12 \pmod{5} \implies\\ k \equiv 4 \pmod{5} \implies\\ k=5j+4$

So, since we have x=3k, we also have x=3(5j+4)=15j+12.

So, clearly j=0 won't work so we should have j=1. That means our money per ticket for the five tickets is:

$\frac{27-12}{5}=3$

And our money per three tickets is:

$\frac{27}{3}=9$

This is easily verifiable. Three tickets needs 27 dollars and 5 tickets needs 15 dollars, which is 12 less than 27 dollars. So we have our money per three dollar ticket at 6 more than money per five dollar.

4 0

1 year ago

Read 2 more answers

In a class, 4/5 of the students have mathematical instruments. 1/4 of these students have lost their protractors. What fraction

Novay_Z [31]

Answer:

$\frac{11}{20}$

Step-by-step explanation:

$\frac{4}{5} - \frac{1}{4}$ <em>Subtract the students who don't have protractors from the students who have mathematical instruments.</em>

$\frac{11}{20}$

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1 year ago

Read 2 more answers

Each of the letters below represents a different digit. if EAT = 721 what does TURKEY represent

zaharov [31]

Thanksgiving

Oh no! Something went wrong while adding your answer
It's too short.<span> Write at least 20 characters to explain it well.</span>

6 0

1 year ago

62% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

velikii [3]

Answer:

a) 0.1829

b) 0.6823

c) 0.0413

Step-by-step explanation:

We are given the following information:

We treat adult having little confidence in the newspaper as a success.

P(Adult have little confidence) = 62% = 0.62

Then the number of adults follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

a) exactly 5

$P(x =5)\\\\=\binom{10}{5}(0.62)^5(1-0.62)^5\\\\=0.1829$

0.1829 is the probability that exactly 5 out of 10 U.S.adults have very little confidence in newspapers.

b) atleast six

$P(x \geq 6)\\\\ = P(x = 6) + P(x = 7)+P(x = 8)+P(x = 9)+P(x=10)\\\\= \displaystyle\sum \binom{10}{n}(0.62)^n(1-0.62)^{10-n}, n =6,7,8,9,10\\\\= 0.6823$

0.6823 is the probability that atleast 6 out of 10 U.S. adults have very little confidence in newspapers.

c) less than four

$P(x$

0.0413 is the probability that less than 4 out of 10 U.S. adults have very little confidence in newspapers.

8 0

1 year ago