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3 years ago

Multiply the square root of 20 and the square root of 63

Mathematics

2 answers:

TEA [102]3 years ago

6 0

Answer:

36.0555

Step-by-step explanation:

harkovskaia [24]3 years ago

4 0

Answer:

= square root of 7 i believe

Step-by-step explanation:

You might be interested in

Is this proportion true 12/15=47/50 16/36=12/27 15/20=24/26 or 4/16=2/14

jekas [21]

For 12/15=47/50, we multiply 12/15 by (1/3)/(1/3) to get 4/5. For 47/50, we multiply it by (1/10)/(1/10) to get the same denominator and 4/7/5. They are not equal.

For 16/36 and 12/27, we'll notice that 36 and 27 are both multiples of 9, so we need to get that as the denominator! Multiply 16/36 by (1/4)/(1/4) because 9 goes into 36 4 times and 12/27 by (1/3)/(1/3) due to that 27 goes into 9 3 times, we get 4/9=4/9 - these are equal!

I challenge you to get the rest of them on your own using my techniques!

3 0

3 years ago

The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp

andreyandreev [35.5K]

Answer:

(a) X ~ N( $\mu=63, \sigma^{2} = 13^{2}$ ).

$\bar X$ ~ N( $\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$ ).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = amount of syrup that people put on their pancakes

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = mean amount of syrup = 63 mL

$\sigma$ = standard deviation = 13 mL

So, the distribution of X ~ N( $\mu=63, \sigma^{2} = 13^{2}$ ).

Let $\bar X$ = sample mean amount of syrup that people put on their pancakes

The z-score probability distribution for the sample mean is given by;

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

where, $\mu$ = mean amount of syrup = 63 mL

$\sigma$ = standard deviation = 13 mL

n = sample of people = 43

So, the distribution of $\bar X$ ~ N( $\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$ ).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X $\leq$ 61.4 mL)

P(X < 62.8 mL) = P( $\frac{X-\mu}{\sigma}$ < $\frac{62.8-63}{13}$ ) = P(Z < -0.02) = 1 - P(Z $\leq$ 0.02)

= 1 - 0.50798 = 0.49202

P(X $\leq$ 61.4 mL) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{61.4-63}{13}$ ) = P(Z $\leq$ -0.12) = 1 - P(Z < 0.12)

= 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < $\bar X$ < 62.8 mL)

P(61.4 mL < $\bar X$ < 62.8 mL) = P( $\bar X$ < 62.8 mL) - P( $\bar X$ $\leq$ 61.4 mL)

P( $\bar X$ < 62.8 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{62.8-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z < -0.10) = 1 - P(Z $\leq$ 0.10)

= 1 - 0.53983 = 0.46017

P( $\bar X$ $\leq$ 61.4 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{61.4-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z $\leq$ -0.81) = 1 - P(Z < 0.81)