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3 years ago

Which condition would prove ΔDEF ∼ ΔJKL?

Mathematics

1 answer:

Svet_ta [14]3 years ago

3 0

Step-by-step explanation:

DE : JK

= 21 : 7

= 1 : 3

<h3>#CMIIW</h3>

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(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop

zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

$\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})$

Compute the sample sizes as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{2\times\hat p(1-\hat p)}{n}$

$n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}$

$=\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787$

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

$\hat p_{1}=0.45\\\hat p_{2}=0.58$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}$

$n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}$

$=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666$

Thus, the sample sizes are 6666.

7 0

3 years ago

Can someone please help me with this!!

denpristay [2]

A^2b^5/c^5 because when you are multiplying exponents you add the powers

3 0

3 years ago

Picture attachment below

Ipatiy [6.2K]

The answer is No, they are not proportional. This is because 8/42 is not equal to 20/105.

6 0

3 years ago

Find the total surface area of the net bellow

9966 [12]

Answer: 102.4 cm^2

Step-by-step explanation:

find the surface area of all of the rectangles

7.2(4)=28.8

7.2(2)=14.4

7.2(4)=28.8

7.2(2)=14.4

4(2)=8

add

28.8+14.4+28.8+14.4+8+8=102.4

4 0

3 years ago

Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age with a mean of 93 seconds and

kkurt [141]

Answer: The running time should at least 119.32 seconds to be in the top 5% of runners.

Step-by-step explanation:

Let X= random variable that represents the running time of men between 18 and 30 years of age.

As per given, X is normally distrusted with mean $\mu=93\text{ seconds}$ and standard deviation $\sigma=16\text{ seconds}$ .

To find: x in top 5% i.e. we need to find x such that P(X<x)=95% or 0.95.

i.e. $P(\dfrac{X-\mu}{\sigma}$

$P(Z$

Since, z-value for 0.95 p-value ( one-tailed) =1.645

So,

Hence, the running time should at least 119.32 seconds to be in the top 5% of runners.

6 0

3 years ago