If the screws that are 4/8 of an inch are laid end to end touching each other how far would the row extend?

In the figure, polygon ABCD is transformed to create polygon A’B’C’D’. This transformation is a

BartSMP [9]

reflection transformation i believe. Correct me if im wrong im not that smart lol.

6 0

2 years ago

In an apartment complex with 28 units, 19 of the renters keep a pet. What percentage does not keep a pet?

Aleks04 [339]

If 19 out of 28 renters keep a pet, there are $28-19 = 9$ renters who don't keep a pet.

Whenever you have a subset of some set, and you want to know which percentage of the set the subset represents, you simply have to compute

$\dfrac{\text{number of elements in the subset}}{\text{number of elements in the set}}\times 100$

So, in your case, you're wondering what percentage of 28 does 9 represent. So, the formula becomes

$\dfrac{9}{28}\times 100 = 0.32\overline{142857}\times 100 = 32.\overline{142857} \approx 32.14\%$

6 0

1 year ago

ΔABC is dilated using a scale factor of 12 to produce ΔA'B'C'. Select all of the statements that apply to the transformation.

GalinKa [24]

Answer:

Options (3) and (6)

Step-by-step explanation:

ΔABC is a dilated using a scale factor of $\frac{1}{2}$ to produce image triangle ΔA'B'C'.

Since, dilation is a rigid transformation,

Angles of both the triangles will be unchanged or congruent.

m∠A = m∠A' and m∠B = m∠B'

Since, sides of ΔA'B'C' = $\frac{1}{2}$ of the sides of ΔABC

Area of ΔA'B'C' = $\frac{1}{2}(\text{Area of triangle ABC})$

Area of ΔABC > Area of ΔA'B'C'

Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.

ΔABC ~ ΔA'B'C'

Therefore, Option (3) and Option (6) are the correct options.

3 0

1 year ago

Two different samples will be taken from the same population of test scores where the population mean and standard deviation are

Alenkinab [10]

Answer:

The sample consisting of 64 data values would give a greater precision.

Step-by-step explanation:

The width of a (1 - α)% confidence interval for population mean μ is:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}$

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (n).

That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.

Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.

The two sample sizes are:

n₁ = 25

n₂ = 64

The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.

Width for n = 25:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{25}}=\frac{1}{5}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]$

Width for n = 64:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]$

Thus, the sample consisting of 64 data values would give a greater precision

5 0

1 year ago

Read 2 more answers

The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar

Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

5 0

1 year ago