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zhenek [66]
3 years ago
5

Which expression completes the equation? 5+7•4 =

Mathematics
1 answer:
abruzzese [7]3 years ago
5 0
5+7*4=48
60÷2+18=48
Answer is D
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(  (\frac{3}{5})  {}^{0} ) {}^{ - 2}  = (1) {}^{ - 2}  =  \frac{1}{1 {}^{2} } =  \frac{1}{1}  = 1

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The inequality describes the range of monthly average temperatures T in degrees Fahrenheit at a certain location. Find an equiva
kolezko [41]

Answer:

Now if the high and low monthly average temperatures satisfy the inequality, then the , monthly averages are always within 22 degrees of 43°F.

Step-by-step explanation:

The inequality describes the range of monthly average temperatures T in degrees Fahrenheit at a certain location.

The inequality expression is given as:

|T-43|\leq 22

now this expression could also be expressed as:

-22\leq T-43\leq22\\\\-22+43 \leq T \leq 22+43\\\\21\leq T\leq 65

Now if the high and low monthly average temperatures satisfy the inequality, then the , monthly averages are always within 22 degrees of 43°F.

( As the difference is 22 degrees to the left and right)

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3 years ago
Statistics question about random probability Cheese pastureized or Raw MilkA cheese can be classified as either raw-milk or past
blsea [12.9K]

Answer:

(a) Two cheeses are chosen at random. the probability that both cheeses are pasteurized is Pr(PP) = 0.82 x 0.82 = 0.6724 to 4 decimal places)

(b) Four cheeses are chosen at random. The probability that all four cheeses are pasteurized is Pr(PPPP) = 0.82 x 0.82 x 0.82 x 0.82 = 0.4521 to 4 decimal places

(c) What is the probability that at least one of four randomly selected cheeses is raw-milk is Pr(RPPP) Or Pr(RRPP) Or Pr(RRRP) Or Pr(RRRR)

= 0.1269 to 4 decimal places

It would not be unusual that at least one of four randomly selected cheeses is raw-milk, because the probability have a value between 0 and 1

Step-by-step explanation:

If is given that 80% of the cheese is classified as pasteurized.

It then implies that 20% of the cheese is classified as Raw-milk

Probability of pasteurized cheese is 0.82(Denoted by Pr(P))

Probability of raw-milk cheese is 0.18(Denoted as Pr(R))

(a) Two cheeses are chosen at random. the probability that both cheeses are pasteurized is Pr(PP) = 0.82 x 0.82 = 0.6724 to 4 decimal places)

(b) Four cheeses are chosen at random. The probability that all four cheeses are pasteurized is Pr(PPPP) = 0.82 x 0.82 x 0.82 x 0.82 = 0.4521 to 4 decimal places

(c) What is the probability that at least one of four randomly selected cheeses is raw-milk is Pr(RPPP) + Pr(RRPP) + Pr(RRRP) + Pr(RRRR)

(0.18 x 0.82 x 0.82 x 0.82) + (0.18 x 0.18 x 0.82 x 0.82) + (0.18 x 0.18 x 0.18 x 0.82) + (0.18 x 0.18 x 0.18 x 0.18) = 0.1269 to 4 decimal places

3 0
3 years ago
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