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2 years ago

50 POINTS: Please help thank you so much!! (Image attached) The polygons are similar. Find the values of the variables.

Mathematics

1 answer:

nadya68 [22]2 years ago

7 0

Answer:

x = 3/5√10

y = 2√10

Step-by-step explanation:

using the Pythagorean theorem:

y² = 7² - 3² = 49 - 9

y² = 40

y = 2√10

3/4.5 = y/5x

4.5y = 15x

substitute for y:

4.5(2√10) = 15x

15x = 9√10

divide both sides of the equation by 15:

x = (9√10)/15 = 3/5√10

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Henry received $500 for his birthday which he intends to invest in an account that pays simple interest at the rate of 12% per y

Mila [183]

Answer:

$560

Step-by-step explanation:

Given that :

Principal, P= $500

Interest rate, r = 12% per year

Amount in account after 1 year

Time = 1 year

Using the relation :

A = P(1 + rt)

A = final amount in account

A = $500(1 + 0.12(1))

A = $500(1 + 0.12)

A = $500(1.12)

A = $560

3 0

2 years ago

Two rectangles are the same one has a length of 11 cm and a width of 10 cm and the other has a width of 7 cm find the length of

ser-zykov [4K]

Answer:

D) 7.7 cm

Step-by-step explanation:

L/W=11cm/10cm=x/7cm

cross multiply to find x

10x=77

divide 10 by both sides

x=7.7

8 0

3 years ago

Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120