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aksik [14]
3 years ago
10

Which statements correctly describe the data shown in the scatter plot?

Mathematics
2 answers:
Doss [256]3 years ago
6 0

Answer:

D

Step-by-step explanation:

Svet_ta [14]3 years ago
5 0

Answer:

Step-by-step explanation:

this will take awhile so imma help u still okay

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Decide whether the triangles are similar. If so, determine the appropriate expression to solve for x.
nikdorinn [45]
Explanation: The given triangles are similar by Angle-Angle postulate, because all three pairs of corresponding angles are congruent.

6 0
3 years ago
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What do I do, please help<br>or just tell me the answer ​
Ket [755]
You can use quizlet or maybe contact a friend
6 0
3 years ago
PLEASE HELP!!!<br> Due in 10 minutes!
erastovalidia [21]

Answer:

41ft²

Step-by-step explanation :

8ft - 2ft = 6ft

6ft - 2ft = 4ft

4ft - 1ft = 3ft

3ft * (8ft + 2ft) / 2 = 15ft²

1ft * (8ft + 2ft) = 10ft²

8ft * 2ft = 16ft²

Total : 16 + 15 + 10 = 41ft²

3 0
2 years ago
Read 2 more answers
The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with a mean of 6.7
hram777 [196]
We need to work out the z-value of each question to work out the probability

Question a)

We have
X = 5 minutes
The mean, μ = 6.7 minutes
Standard deviation, σ = 2.2 minutes
z-score = (X - μ) / σ = (5 - 6.7) / 2.2 = -0.77

We want to find the probability of z < -0.77 so we read the z-table for the value of z on the left of -0.77 we have the probability P(z< -0.77) = 0.2206

The table is attached in picture 1 below

The probability of assembly time less than 5 minutes is 0.2206 = 22.06%

Question b)

z-score = (10-5) / 2.2 = 2.27

Reading the z-table for P(z<2.27) = 0.9884
The table reading is shown in the second picture below

The probability the assembly time will be less than 10 minutes is 0.9884

We can use this information to find the probability of the assembly time will be more than 10 minutes  = 1 - 0.9884 = 0.0116 = 1.16%

Question c)

The value of z between 5 minutes and 10 minutes is

P(-0.77<z<2.27) = P(z<2.27) - P(z< -0.77)
P(-0.77<z<2.27) = 0.9884 - 0.2206 = 0.7678 = 76.78%


6 0
3 years ago
A random sample of 85 group leaders, supervisors, and similar personnel revealed that a person spent an average 6.5 years on the
strojnjashka [21]

Answer:

<em>95% of confidence interval for the Population</em>

<em>( 6.1386 , 6.8614)</em>

Step-by-step explanation:

<u><em>Step( i ):-</em></u>

<em>Given random sample size 'n' =85</em>

<em>Mean of the sample size x⁻ = 6.5 years</em>

<em>Standard deviation of Population = 1.7 years</em>

<em>Level of significance = 0.95 or 0.05</em>

<u><em>Step(ii):-</em></u>

<em>95% of confidence interval for the Population is determined by</em>

<em></em>(x^{-} - Z_{0.05} \frac{S.D}{\sqrt{n} } , x^{-} + Z_{0.05} \frac{S.D}{\sqrt{n} })<em></em>

<em></em>(6.5 - 1.96\frac{1.7}{\sqrt{85} } , 6.5 + 1.96 \frac{1.7}{\sqrt{85} })<em></em>

<em>( 6.5 - 0.3614 , 6.5 + 0.3614 )</em>

<em>( 6.1386 , 6.8614)</em>

<u><em>Conclusion</em></u><em>:-</em>

<em>95% of confidence interval for the Population</em>

<em>( 6.1386 , 6.8614)</em>

8 0
4 years ago
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