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3 years ago

Express 4 8/9 as an improper fraction A 36/9 B 21/9 C 44/9 D 40/9

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

8 0

The answer is c because 9×4=36 then add eight which equals 44. therefore, it is 44/9

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How many passwords of 4 digits are there(a)if none of the digits can be repeated?(b)start with 5 and end in an even digit?

liraira [26]

Answer:

5040,56

Step-by-step explanation:

We have to construct pass words of 4 digits

a) None of the digits can be repeated

We have total digits as 0 to 9.

4 digits can be selected form these 10 in 10P4 ways (since order matters in numbers)

No of passwords = 10P4

= $10(9)(8)(7)\\=5040$

b) start with 5 and end in even digit

Here we restrain the choices by putting conditions

I digit is compulsorily 5 and hence only one way

Last digit can be any one of 0,2,4,6,8 hence 5 ways

Once first and last selected remaining 2 digits can be selected from remaining 8 digits in 8P2 ways (order counts here)

=56

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3 years ago

What is the scale factor of 5mm:1cm

Goryan [66]

<span>
1 cm = 10mm

5 mm : 10mm : :1 : 2

hoped i helped. ^-^
</span>

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3 years ago

Please help me!! Just help me solve

Kruka [31]

The solution for the system of equations are:
$x = \frac{7}{5} \\ y = \frac{11}{5}$
Attached is the graph of lines y=3x-2 and y=-2x+5.

4 0

3 years ago

One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo

almond37 [142]

Answer:

(1) P( $\bar X$ < 26 inches) = 0.0436

(2) P(27.5 inches < $\bar X$ < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let $\bar X$ = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

Z = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean vertical leap = 28 inches

$\sigma$ = standard deviation = 7 inches

n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P( $\bar X$ < 26 inches)

P( $\bar X$ < 26) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{26-28}{\frac{7}{\sqrt{36} } }$ ) = P(Z < -1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < $\bar X$ < 28.5 inches) = P( $\bar X$ < 28.5 inches) - P( $\bar X$ $\leq$ 27.5 inches)

P( $\bar X$ < 28.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{28.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z < 0.36) = 0.64058 {using z table}

P( $\bar X$ $\leq$ 27.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{27.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z $\leq$ -0.36) = 1 - P(Z < 0.36)

= 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < $\bar X$ < 28.5 inches) = 0.64058 - 0.35942 = 0.2812

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3 years ago

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Elena L [17]

Answer is B...........

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3 years ago