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3 years ago

If you run 15 laps per practice, how many practices will it take you to run 30 laps?

2 answers:

6 0

First you divide 30 by 15 because you are finding out how many practices it takes to run 30 laps.

30 divided by 15 = 2

It will take 2 laps to run 30 laps.

Answer: 2

NNADVOKAT [17]3 years ago

6 0

Answer:

1 round = 15 laps

? round = 30 laps

30 / 15 × 1 = 2 rounds

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What is the slope of the line conaning the points (-3,4)and(2,1)

olya-2409 [2.1K]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -3 &,& 4~) 
% (c,d)
&&(~ 2 &,& 1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-4}{2-(-3)}\implies \cfrac{1-4}{2+3}\implies \cfrac{-3}{5}$

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3 years ago

A ball is thrown from a height of 45 meters with an initial downward velocity of 10 m/s. The ball's height h (in meters) after t

dmitriy555 [2]

Answer:

2.16 s

Step-by-step explanation:

When it hits the ground, height h=0 hence

$0=45-10t-5t^{2}$

$5t^{2}+10t-45=0$

$t^{2}+2t-9=0$

Using quadratic formula

$x=-1+\sqrt 10\\ x=-1-\sqrt 10$

Since time can't be negative, $-1+\sqrt 10$

t=2.16 s

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3 years ago

How many terms are in the given expression? 52 + Zn - 2f

tatuchka [14]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

In the given figure poq is a line. if x=30 then find qor and ros

LuckyWell [14K]

Answer:

$2y + 3y + x = 180 \\ 5y + 30 = 180 \\ 5y = 180 - 30 \\ 5y = 150 \\ y = 30$

QOR =3Y

=3×30

=90°

ROS = 2Y

=2×30

=60°

3 0

2 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)

= 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

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

4 0

3 years ago