Can someone please help me with this

Y=3x-1 identify the slope and y-intercerpt

soldier1979 [14.2K]

Answer: The slope is 3 and the y-intercept is -1

Step-by-step explanation: This is right bro. 100%

7 0

3 years ago

Read 2 more answers

Henry has 20 Lawns mowed out of a total of 50 lawns. What percentage of the lawns does Henry have to still mow?

elena55 [62]

Henry has completed 40% of his work, leaving him with a remaining 60%.

What you do is divide 20 by 50 and your answer is .4, move the decimal two places to the right to get your percent, 40. Then subtract it out of 100 and you get the remaining 60%, which essentially is your answer.

5 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)