Line AB contains points A(-6,3) and B(2,-5). Line AB has a slope that is?
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Using the z-table, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.
For a normally distributed set of data, given the mean and standard deviation, the probability can be determined by solving the z-score and using the z-table.
First, solve for the z-score using the formula below.
z-score = (x – μ) / σ
where x = individual data value = 100
μ = mean = 125
σ = standard deviation = 18
z-score = (100 – 125) / 18
z-score = (-25) / 18
z-score = -1.39
Find the probability that corresponds to the z-score in the z-table. (see attached images)
-1.39 - (-1.3) : -1.4 - (-1.3) = x - 0.0968 : 0.0808 - 0.0968
-0.09 : -0.1 = x - 0.0968 : -0.016
x - 0.0968 = -0.09(-0.016)/-0.1
x = -0.0144 + 0.0968
x = 0.0824
x = 0.0824
Hence, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.
Learn more about probability here: brainly.com/question/26822684
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To find the z-score for a weight of 196 oz., use
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.