Negative 7 divided by 14/3 as a fraction

What is the slope-intercept form of the linear equation 2x–8y=32?

skelet666 [1.2K]

Answer: y = 1/4x - 4

Step-by-step explanation: write it in slope intercept form y = mx + b

3 0

3 years ago

How do you do it????

m_a_m_a [10]

Answer:find the slope

Step-by-step explanation:

I don't know if you know how to find the slope if not reply back. But that's what the equation wants you to do.

6 0

3 years ago

Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation o

hjlf

Answer:

46.18% of the items will weigh between 6.4 and 8.9 ounces.

Step-by-step explanation:

We are given that the weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.

Let X = weight of items produced by a machine

The z-score probability distribution for is given by;

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

where, $\mu$ = mean weight = 8 ounces

$\sigma$ = standard deviation = 2 ounces

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percentage of items that will weigh between 6.4 and 8.9 ounces is given by = P(6.4 < X < 8.9) = P(X < 8.9 ounces) - P(X $\leq$ 6.4 ounces)

P(X < 8.9) = P( $\frac{ X -\mu}{\sigma}$ < $\frac{ 8.9-8}{2}$ ) = P(Z < 0.45) = 0.67364 {using z table}

P(X $\leq$ 70) = P( $\frac{ X -\mu}{\sigma}$ $\leq$ $\frac{ 6.4-8}{2}$ ) = P(Z $\leq$ -0.80) = 1 - P(Z < 0.80)

= 1 - 0.78814 = 0.21186

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.45 and x = 0.80 in the z table which has an area of 0.67364 and 0.78814 respectively.

Therefore, P(6.4 < X < 8.9) = 0.67364 - 0.21186 = 0.4618 or 46.18%

Hence, 46.18% of the items will weigh between 6.4 and 8.9 ounces.

8 0

3 years ago

Can someone help me with this problem? am i right? this is 7th grade btw.

labwork [276]

I think you are right, it is (1,1).

6 0

3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

e) $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago