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3 years ago

Write the smallest possible length. length:9.15 (to 2 decimal places)

Mathematics

2 answers:

kicyunya [14]3 years ago

6 0

Answer:

The smallest possible length of 9.15 rounded to 2 decimal places is 9.145

Step-by-step explanation:

When you round a number to 2 decimal places you round the 3rd decimal place to the nearest integer, then numbers 9.140, 9.141, 9.142, 9.143 and 9.144 are rounded to 9.14, and numbers 9.145, 9.146, 9.147, 9.148 and 9.149 are rounded to 9.15.

pentagon [3]3 years ago

3 0

<span>9.145 is the smallest value that will round to 9.15.

</span><span>As the rule that if that digit is less than 5, do not change the rounding digit but drop all digits to the right of it. So it should be 5 to 9 but then the question above is looking for the smallest possible number which is 5. </span>

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Suppose that a certain movie earns $3,000,000 in ticket sales. If 60% goes to the movie and 40% goes to theatres: Movie studios

miskamm [114]

Answer: See explanation

Step-by-step explanation:

Amount earned by movie = $3,000,000.

Percentage that went to movie = 60%

Percentage that went to theater = 40%

Revenue gotten by movie studio = 60% × $3,000,000

= 0.6 × $3,000,000

= $1,800,000

Revenue gotten by theaters = 40% × $3,000,000

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3 years ago

2qt/min=how many gal/s

Scrat [10]

Answer:

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8 0

3 years ago

Read 2 more answers

A track star runs two races on a certain day. The probability thathe wins the first race is 0.7, the probability that he wins th

NARA [144]

Answer:

a) 80% probability that he wins at least one race.

b) 30% probability that he wins exactly one race.

c) 20% probability that he wins neither race.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that he wins the first race.

B is the probability that he wins the second race.

C is the probability that he does not win any of these races.

We have that:

$A = a + (A \cap B)$

In which a is the probability that he wins the first race but not the second and $A \cap B$ is the probability that he wins both these races.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that he wins both races is 0.5.

This means that $A \cap B = 0.5$

The probability that he wins the second race is 0.6

This means that $B = 0.6$

$B = b + (A \cap B)$

$0.6 = b + 0.5$

$b = 0.1$

The probability that he wins the first race is 0.7.

This means that $A = 0.6$

$A = a + (A \cap B)$

$0.7 = a + 0.5$

$a = 0.2$

A) he wins at least one race.

This is

$P = a + b + (A \cap B) = 0.2 + 0.1 + 0.5 = 0.8$

There is an 80% probability that he wins at least one race.

B) he wins exactly one race.

This is

$P = a + b = 0.2 + 0.1 = 0.3$

There is a 30% probability that he wins exactly one race.

C) he wins neither race

Either he wins at least one race, or he wins neither. The sum of these probabilities is 100%.

From a), we have that there is an 80% probability that he wins at least one race.

So there is a 100-80 = 20% probability that he wins neither race.

6 0

3 years ago

Can someone please help me with math?

Fofino [41]

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3 years ago

7 to the negative 3rd porwer times 7 to the negative 2nd porwer times 2 the 0 power divided by 7 to negative 1st power times 7 t

Jobisdone [24]

oof

remember some rules

(a/b)(c/d)=(ac)/(bd)

$\frac{x^a}{x^b}=x^{a-b}$

$x^0=1$

$(x^a)(x^b)=x^{a+b}$

$x^{-a}=\frac{1}{x^a}$

$\frac{(7^{-3})(7^{-2})(2^0)}{(7^{-1})(7^{-4})(2^3)}=$

$\frac{(7^{-3-2})(2^0)}{(7^{-1-4})(2^3)}=$

$\frac{(7^{-5})(2^0)}{(7^{-5})(2^3)}=$

$(\frac{7^{-5}}{7^{-5}})(\frac{2^0}{2^3})=$

$(7^{-5-(-5)})(2^{0-3})=$

$(7^0)(2^{-3})=$

$\frac{1}{2^3}=$

$\frac{1}{8}$

7 0

3 years ago