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3 years ago

5. On a recent day, the exchange rate for US dollars to British pounds

Mathematics

1 answer:

attashe74 [19]3 years ago

8 0

250*0.62= 155

155 pounds is your answer.

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Simplify in Exponential form First correct answer gets BRAINLIEST

svlad2 [7]

Answer:

Steps to simplifying fractions

Therefore, 9/9 simplified to lowest terms is 1/1.

Reduce 9/8 to lowest terms

9/8 is already in the simplest form. It can be written as 1.125 in decimal form (rounded to 6 decimal places).Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In the diagram below AFB is congruent to EFD if EFD = 5x + 6, DFC = 19x - 15, and EFC = 17x + 19, find AFE

yanalaym [24]

The angle m∠AFE is 128 degrees.

<h3>How to find angles?</h3>

∠AFB ≅ ∠EFD

∠EFD = 5x + 6

m∠DFC = (19x - 15)°

m∠EFC = (17x + 19)°

m∠AFE = ?

m∠AFB + m ∠EFD + m∠AFE = 180

Therefore,

5x + 6 + 5x + 6 + m∠AFE = 180

5x + 5x + 6 + 6 + m∠AFE = 180

10x + 12 + m∠AFE = 180

10x + m∠AFE = 180 - 12

10x + m∠AFE = 168

m∠AFE = 168 - 10x

m∠EFC = m ∠EFD + m∠DFC

17x + 19 = 5x + 6 + 19x - 15

17x - 5x - 19x = 6 - 15 - 19

-7x = - 28

x = 28 / 7

x = 4

Therefore,

m∠AFE = 168 - 10x

m∠AFE = 168 - 10(4)

m∠AFE = 168 - 40

m∠AFE = 128°

Therefore, the angle m∠AFE = 128°

learn more on angles here: brainly.com/question/13212279

#SPJ1

7 0

2 years ago

Radical expressions Help pls multiple choice

Ann [662]

Answer:

4th option (sqrt(10))/2

Step-by-step explanation:

$\frac{5}{\sqrt{10} } * \frac{\sqrt{10}}{\sqrt{10} } \\ \frac{5\sqrt{10}}{10} \\\\frac{\sqrt{10}}{2}$

8 0

3 years ago

If AB=BC, AB=4x-2, and BC= 3x+3, find AB

a_sh-v [17]

Isnt it 4x-2 ?? i mean it does say "find" AB...

6 0

3 years ago

How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide

kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

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

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

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

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The information provided is:

σ = $60

MOE = $2

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}$

$=[\frac{1.96\times 60}{2}]^{2}$

$=3457.44\\\approx 3458$

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0

3 years ago