Really stuck on this question

Find value of x such that the data set has a mean of 105: 99, 119, 106, 112; 108

Marizza181 [45]

I'm assuming that x is part of the data set, and, with x, the mean equals 105. To find the value of x, you must add all the data values together to get 544+x (you still don't know what x equals). Then put 544+x over how many days values there are, including x (there are 6). You should have 544+x/6. Now, as this is how you would calculate the mean if you knew what x was equal to, you must set it equal to the mean, since you know what it is (105). You should now have 544+x/6 = 105. You have your equation set up--now you just have to solve it. I would multiply by 6 on both sides to get rid of the 6 on the left side. You would then have 544+x = 630. I would finally subtract 544 from both sides to get x = 86. Your final answer is x = 86.

6 0

3 years ago

Which diagram represents the vector addition C=A+B. Which represents vector subtraction C=A-B?

olga55 [171]

If the vectors point towards the same direction, the magnitude of the vector quantities are added as in, C = A + B. If they point towards opposite direction, the magnitudes are subtracted giving a resultant vector equal to C = A - B.

3 0

3 years ago

The perimeter of the blueprint is represented by the equation P=8x. What is the equation of the perimeter when solved for x

Korvikt [17]

Answer:

x = P/8

Step-by-step explanation:

The perimeter of the blueprint is given by the equation :

P = 8x

We need to solve the above equation for x.

Dividing both sides of the equation by 8.

$\dfrac{P}{8}=\dfrac{8x}{8}\\\\x=\dfrac{P}{8}$

Hence, the value of x is P/8

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)

= 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

The ratio of the number of fiction book is 9:4, the different between the number of fiction and non fiction is 245

Evgesh-ka [11]

Answer:

Step-by-step explanation:

The ratio of the number of fiction books is 9:4

the difference between the number of fiction and nonfictionn is 245

so if the number of fiction books = 9x

then the number of nonfiction books = 4x

then difference = 9x-4x = 5x

the difference is given 245

then 5x= 245

x= $\frac{245}{5}$ =49

number of fiction books =49×9=441

number of nonfiction books =49×4=196

brainly.com/question/27979540?answering=true&answeringSource=feedPublic%2FhomePage%2F12

#SPJ2

8 0

2 years ago

Really stuck on this question​

Really stuck on this question