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GalinKa [24]
2 years ago
9

Find the width of the river in Figure 20.18 to the nearest metre.​

Mathematics
1 answer:
Gnom [1K]2 years ago
5 0

Answer:

~ 6 m

Step-by-step explanation:

For a right triangle

tan = opposite leg / adjacent leg

     = height / width of river

     

tan 15 = 1.6 / width

width = 1.6 / tan 15 = 5.97 m

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(5x^{2}+3x-1)-(3x^{2}-4x-6)<br> find the difference PLEASE HURRY IM TIMED
Andreyy89

Answer:

=2x2+7x+5

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
An exam starts at 10:10 and lasts for 75 minutes what time does the exam end pls help i will give brainliest
Sidana [21]

Given:

An exam starts at 10:10 and lasts for 75 minutes.

To find:

The time when the exam end.

Solution:

We have,

Starting time = 10:10

Exam duration = 75 minutes

First we will divide the 75 minutes in two parts 50 minutes and 25 minutes.

We know that 50 minutes after 10:10 is 11:00 and 25 minutes after 11:00 is 11:25. So,

10:10 + 75 minutes = 10:10 + 50 minutes +25 minutes)

                               = (10:10 + 50 minutes) +25 minutes

                               = 11:00 +25 minutes

                               = 11:25

Therefore, the exam ends at 11:25.

3 0
3 years ago
Compute the standard error for sample proportions from a population with proportion p= 0.55 for sample sizes of n=30, n=100 and
Pani-rosa [81]

Given Information:

Population proportion = p =  0.55

Sample size 1 = n₁ = 30

Sample size 2 = n₂ = 100

Sample size 3 = n₃ = 1000

Required Information:

Standard error = σ = ?

Answer:

$ \sigma_1 = 0.091 $

$ \sigma_2 = 0.050 $

$ \sigma_3 = 0.016 $

Step-by-step explanation:

The standard error for sample proportions from a population is given by

$ \sigma =  \sqrt{\frac{p(1-p)}{n} } $  

Where p is the population proportion and n is the sample size.

For sample size n₁ = 30

$ \sigma_1 =  \sqrt{\frac{p(1-p)}{n_1} } $

$ \sigma_1 =  \sqrt{\frac{0.55(1-0.55)}{30} } $

$ \sigma_1 = 0.091 $

For sample size n₂ = 100

$ \sigma_2 =  \sqrt{\frac{p(1-p)}{n_2} } $

$ \sigma_2 =  \sqrt{\frac{0.55(1-0.55)}{100} } $

$ \sigma_2 = 0.050 $

For sample size n₃ = 1000

$ \sigma_3 =  \sqrt{\frac{p(1-p)}{n_3} } $

$ \sigma_3 =  \sqrt{\frac{0.55(1-0.55)}{1000} } $

$ \sigma_3 = 0.016 $

As you can notice, the standard error decreases as the sample size increases.

Therefore, the greater the sample size lesser will be the standard error.

8 0
4 years ago
Simplify 5(cis7pi/4)^4
solniwko [45]

Answer:

  -625

Step-by-step explanation:

The magnitude gets raised to the power, and the angle gets multiplied by the power:

  5(cis7pi/4)^4 = (5^4)cis(4·7π/4) = 625cis(7π) = 625cis(π) = -625

5 0
3 years ago
How do I find the answer to the second question?
dimaraw [331]
The sentence does not make sense because a test can only be 100% not 200

7 0
4 years ago
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