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worty [1.4K]
1 year ago
12

The conversion rate for US Dollars to Costa Rican Colones is 1 USD = 518 Colones.

Mathematics
1 answer:
allsm [11]1 year ago
5 0

The cost of 0.5 kg of bananas is 393.60 Colones as per the given conversion rates

Conversion rate of 1 USD to Costa Rican Colones = 518 Colones

The conversion rate of kg to pounds given in the question: 1 kg = 2.2025 lbs

Cost of one pound of bananas = $0.69

Bananas required to be purchased = 0.5kg

Converting 0.5kg bananas to pounds = 0.5*2.2025 = 1.10125 pounds

Cost of 1.10125 pound of bananas in dollars = 1.10125*0.69 = 0.7598

Cost of 1.1025 pounds of bananas in Colones = 0.7598*518 = 393.60 Colones

Hence, the cost is 393.60

Therefore, the cost of 0.5 kg bananas in Colones is 393.60 Colones

Learn more about conversion rate:

brainly.com/question/2274822

#SPJ1

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Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.
olasank [31]

Answer:

x=18 y=69

Step-by-step explanation:

3x+5y=399

x+5y=363

2x=36

x=18

18+5y=363

5y=345

y=69

5 0
2 years ago
William made 1/2 gallon of lemonade to divide equally among 3 people. How much lemonade will each person?
katovenus [111]
<span>a) 1/6 gallon all u need to do is divide</span>
8 0
3 years ago
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Consider independent simple random samples that are taken to test the difference between the means of two populations. The varia
Arturiano [62]

Answer:

d. t distribution with df = 80

Step-by-step explanation:

Assuming this problem:

Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown, but are assumed to be equal. The sample sizes of each population are n1 = 37 and n2 = 45. The appropriate distribution to use is the:

a. t distribution with df = 82.

b. t distribution with df = 81.

c. t distribution with df = 41.

d. t distribution with df = 80

Solution to the problem

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

This last one is an unbiased estimator of the common variance \sigma^2

So on this case the degrees of freedom are given by:

df= 37+45-2=80

And the best answer is:

d. t distribution with df = 80

5 0
3 years ago
Claire jogs 6 miles per hour and Megan jogs 5 miles per hour. they start together at their campsite and jog to an outpost 18 mil
horrorfan [7]

Answer:

16.364 miles

Step-by-step explanation:

Speed of Claire = 6 miles per hour

Speed of Megan= 5 miles per hour

Distance from camp site to out post = 18 miles

Time for Claire to travel=18/6= 3 hours

Time for Megan to travel = 18/5 = 3.6 hours

The time they meet will be equal

Claire distance= 18 + x

Megan distance = 18-x

Claire time =( 18+x)/6

Megan Tim =( 18-x)/5

(18+x)/6 =(18-x)/5

5(18+x)= 6(18-x)

90 +5x = 108-6x

11x=18

X=18/11

X=1.636 Miles

So the distance from the Outpost to where they meet = 18-1.636

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6 0
3 years ago
Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio
adoni [48]

Answer:

a : b = 4 : 3

Step-by-step explanation:

(mark as brainlist answer)

According to the given condition is the question :

Two numbers a and b are such that the sum of 5% of a and 4% of b is two-third of the sum of 6% of a and 8% of b

⇒ (5% of a) + (4% of b) = (2/3) × [(6% of a) + (8% of b)

⇒ 0.05a + 0.04b = 2/3 × (0.06a + 0.08b)

⇒ 0.15a + 0.12b = 0.12a + 0.16b

⇒ 0.03a = 0.04b

⇒ a/b = 4/3

⇒ a : b = 4 : 3

8 0
3 years ago
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