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4 years ago

Kenya exchanges $200 for euros (€). Suppose the conversion rate is €1 = $1.321. How many euros should Kenya receive?

Mathematics

2 answers:

zzz [600]4 years ago

8 0

Answer:

The second option, or option b

Step-by-step explanation:

200/1.321 = 151.4

Fed [463]4 years ago

4 0

Answer: € 151.40

Step-by-step explanation:

€1 = $1.321

200/1.321 = €151.40

You might be interested in

Solve for x. Assume that lines which appear to be tangent are tangent.

stiv31 [10]

Answer:

x = 1

Step-by-step explanation:

In order to solve the given problem, we should have the knowledge of the following property.

Measure of the tangents drawn from external points to a circle are equal.

46x - 1 = 45x

46 x - 45x = 1

x = 1

4 0

3 years ago

Select the simplification that accurately explains the following statement.

lutik1710 [3]

Answer:

Option (b) is correct.

$(2^\frac{1}{4} )^4=2^\frac{1}{4} \times 2^\frac{1}{4}\times 2^\frac{1}{4}\times 2^\frac{1}{4}=2^{(\frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4} )}=2^{1}=2$

Step-by-step explanation:

Given: $(2^\frac{1}{4} )^4$

We have too choose the correct simplification for the given statement.

Consider $(2^\frac{1}{4} )^4$

Using property of exponents, $(a^m)^n=a^m\times a^m\times a^m\times ....\times (n\ times)$

We have,

$(2^\frac{1}{4} )^4=2^\frac{1}{4} \times 2^\frac{1}{4}\times 2^\frac{1}{4}\times 2^\frac{1}{4}$

Again applying property of exponents $a^m\times a^m=a^{n+m}$

We have,

$(2^\frac{1}{4} )^4=2^{(\frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4} )}$

Simplify, we have,

$(2^\frac{1}{4} )^4=2^{\frac{4}{4}}$

we get,

$(2^\frac{1}{4} )^4=2^{1}=2$

Thus, $(2^\frac{1}{4} )^4=2$

Option (b) is correct.

8 0

3 years ago

Read 2 more answers

I really need help on these two. Its fine if you dont get both of them. i will give you 25 points

masya89 [10]

Answer:

He read 1/3 of the book on the second day

He didn't finish the book in two days.

Step-by-step explanation:

Lets find 3/5 of 5/9

If we said, find 3/5 of x, the answer would be 3/5x.

3/5*x = 3/5x

In the same way, $\frac{3}{5} *\frac{5}{9} = \frac{15}{45} = \frac{1}{3}$

He read 1/3 of the book on the second day

Does 5/9 + 1/3 equal to 1?

1/3 = 3/9

5/9 + 3/9 = 8/9

He didn't finish the book in two days.

Hope this helps :)

Have a nice day!

3 0

3 years ago

A car is driving away from a crosswalk. The formula d = t2 + 2t expresses the car's distance from the crosswalk in feet, d, in t

alukav5142 [94]

Answer:

8 feet per second.

Step-by-step explanation:

We have been given that a car is driving away from a crosswalk. The formula $d=t^2+2t$ expresses the car's distance from the crosswalk in feet, d, in terms of the number of seconds, t, since the car started moving.

We will use average change formula to solve our given problem.

$\text{Average change}=\frac{f(b)-f(a)}{b-a}$

$\text{Average change}=\frac{d(5)-d(1)}{5-1}$

$\text{Average change}=\frac{(5)^2+2(5)-((1)^2+2(1))}{4}$

$\text{Average change}=\frac{25+10-(1+2)}{4}$

$\text{Average change}=\frac{35-3}{4}\\\\\text{Average change}=\frac{32}{4}\\\\\text{Average change}=8$

Therefore, the the car's average speed over the given interval of time would be 8 feet per second.

8 0

3 years ago

We are interested in the amount that students study per week. Suppose you collected the following data in hours {4.4, 5.2, 6.4,

olga nikolaevna [1]

Answer:

Step-by-step explanation:

Hello!

The objective is to estimate the average time a student studies per week.

A sample of 8 students was taken and the time they spent studying in one week was recorded.

4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4

n= 8

X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74

S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94

S= 1.39

Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:

X[bar] ± $t_{n-1;1-\alpha /2}$ * (S/√n)

$t_{n-1;1-\alpha /2}= t_{7;0.975}= 2.365$

6.74 ± 2.365 * (1.36/√8)

[5.6;7.88]

Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]

I hope this helps!

3 0

4 years ago

Read 2 more answers