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Lesechka [4]
3 years ago
15

15. B(-2, -9), C(0,-5), D(6,-3), E(4, -7)

Mathematics
1 answer:
artcher [175]3 years ago
5 0

Answer:

BC= 2√5

CD= 2√10

DE= 2√5

BE= 2√10

slope of BC = 2

slope of CD = 1/3

slope of DE = 2

slope of BE = 1/3

Step-by-step explanation:

formula for distance = √(x2-x1)²+(y2-y1)²

formula for slope = (y2-y1)/(x2-x1)

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Polygons, Number of sides, Measure is angles, and angle sums
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3 years ago
40% of oatypop cereal boxes contain a prize Hannah plans to keep buying cereal until she gets the prize. What is the probability
AveGali [126]

Answer:

86%

Step-by-step explanation:

70+37+22=127

129/150= 0.86=86%

7 0
4 years ago
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As quality control manager at a raisin manufacturing and packaging plant, you want to ensure that all the boxes of raisins you s
9966 [12]

Answer:

Step-by-step explanation:

The population average number of raisins per box is 30.

The type of statistical analysis to be run here is the inferential statistical analysis and its assumptions include normality, linearity and equality of variance.

Null hypothesis: Average number of raisins per box is 30: u = 30

Alternative hypothesis: Average number of raisins per box is not 30: u ≠ 30

To perform a two tailed single sample test, we have

z statistic = (the sample mean - hypothesized mean) / (SD/√n)

sample mean = 28.05, SD = 3.87 hypothesized mean = 30 n = 36

Thus, we have 28.05 - 30 / (3.87/√36)

= -1.95 / (3.87/6)

= -1.95 / 0.645

= -3.023

we say this is a left tailed test and use the negative z table

where -3.02 = 0.0013

To determine whether the average number of raisins per box differs from the expected 30.

Using p < 0.05, since the observed value 0.0013 is less than 0.05, we conclude the the average number of raisins per box differs from the expected 30.

6 0
4 years ago
The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial
xxTIMURxx [149]

Step-by-step explanation:

Let the initial population of a community be P0 and the population after time t is P(t).

If the population of a community is known to increase at a rate proportional to the number of people present at time t, this is expressed as:

P(t) = P_0e^{kt}

at t = 5 years, P(t) = 2P0

substitute:

2P_0 = P_0e^{5k}\\2 = e^{5k}\\ln2 = lne^{5k}\\ln2 = 5k\\k = \frac{ln2}{5}\\k = 0.1386

If the population is 9,000 after 3 years

at t = 3, P(t) = 9000

a) Substitute into the formula to get P0

9000 = P_0e^{0.1386\times 3}\\9000 = P_0e^{0.4158}\\9000 = 1.5156P_0\\P_0 = \frac{9000}{1.5156}\\ P_0 = 5938.24

Hence the initial population is approximately 5938.

b) In order to know how fast the population growing at t = 10, we will substitute t = 10 into the formula as shown:

P(10) = 5938.24e^{0.1386(10)}\\P(10) = 5938.24e^{1.386}\\P(10) = 5938.24(3.9988)\\P(10) = 23,745.97

Hence the population of the community after 10 years is approximately 23,746

4 0
4 years ago
How does the variable of data set A compare to the variability of data set B.
cestrela7 [59]
The measures of variability for each class are listed:

Class A:                         Class B:
Range : 30                     Range: 30
IQR: 12.5                       IQR: 20.5
MAD: 7.2                       MAD: 9.2

The interquartile range and the mean absolute deviations for class B are larger number therefore indicating that there is more variability in that data set.
4 0
3 years ago
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