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raketka [301]
2 years ago
11

Determine whether the polygons are similar. If so, identify the correct similarity ratio and the similarity statement. HELP ASAP

!!

Mathematics
1 answer:
Vilka [71]2 years ago
7 0

Answer:

  No, the triangles are not similar

Step-by-step explanation:

The (reduced) side length ratios, shortest to longest, are ...

  12 : 18 : 20 = 6 : 9 : 10

and

  5 : 12 : 13

These are not the same, so the triangles are not similar.

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bezimeni [28]

Answer:

y=-3x+7

Step-by-step explanation:

m=gradient=-3

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2 years ago
Hi someone please solve this
pshichka [43]

Answer:

x = 115

Step-by-step explanation:

If lines are parallel, alternate interior angles are congruent (the same measure). Since this is a parallelogram, opposite sides are parallel, and AIA is applicable, so x = 115 degrees.

6 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.6 c
mafiozo [28]

Answer:

(a) the probability that an individual distance is greater than 214.80 cm is 0.1401.

(b) The probability that the mean for 15 randomly selected distances is greater than 204.00 cm is 0.2482.

(c) The normal distribution can be used because the original population has a normal distribution.

Step-by-step explanation:

We are given that the overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.6 cm.

(a) Let X = <em>the overhead reach distances of adult females</em>.

The z-score probability distribution for the normal distribution is given by;

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

where, \mu = population mean reach distance = 205.5 cm  

           \sigma = standard deviation = 8.6 cm

So, X ~ Normal(\mu=205.5,\sigma^{2} =8.6^{2})

Now, the probability that an individual distance is greater than 214.80 cm is given by = P(X > 214.80 cm)

    P(X > 214.80 cm) = P( \frac{X-\mu}{\sigma} > \frac{214.80-205.5}{8.6} ) = P(Z > 1.08) = 1 - P(Z \leq 1.08)

                                                                    = 1 - 0.8599 = <u>0.1401</u>

The above probability is calculated by looking at the value of x = 1.08 in the z table which has an area of 0.8599.

(b) Let \bar X = <em>the sample mean selected distances</em>.

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 = population mean reach distance = 205.5 cm  

           \sigma = standard deviation = 8.6 cm

           n = sample size = 15

Now, the probability that the mean for 15 randomly selected distances is greater than 204.00 cm is given by = P(\bar X > 204.00 cm)

    P(\bar X > 204 cm) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{204-205.5}{\frac{8.6}{\sqrt{15} } } ) = P(Z > -0.68) = 1 - P(Z \leq 0.68)

                                                                    = 1 - 0.7518 = <u>0.2482</u>

The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.7518.

(c) The normal distribution can be used in part​ (b), even though the sample size does not exceed​ 30 because the original population has a normal distribution and the sample of 15 randomly selected distances has been taken from the population itself.

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3 years ago
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Lynna [10]
I hope this helps you

7 0
3 years ago
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The table shows the result of a poll of 150 randomly selected middle school students who were asked if they take French or Spani
seraphim [82]

Answer:

25/37

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Number of 7th graders offering french= 25

Total number of 7 graders = 37

P( 7th Grade| French) = 25/37

5 0
2 years ago
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