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2 years ago

9

in a bag of candy, there are 6 blue candies, 5 yellow candies, 1 red candy, and 12 green candies. write the following ratios. A.

The number of red candies to the number of blue candies

1 answer:

goldenfox [79]2 years ago

3 0

Answer:

1:6

Step-by-step explanation:

1 red candy to 6 blue candy.

In other words, for every 6 blue candies there will be 1 red candy.

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0.05 is 1/10 the value of <br>A. 5 <br>B. 0.005<br>C. 0.5 <br>D. 50<br>

anastassius [24]

Answer:

C.

Step-by-step explanation:

as 0.05 is 1/10 of a number, multiplying 0.05 by 10 will give you the number that when divided by 10 will give you 0.05. Hence just multiply 0.05 by 10 and you get 0.5.

8 0

3 years ago

Which shows the decimals 34.8, 36.43, and 36.29 in order from greatest to least?

ruslelena [56]

All you do is add a zero on the end of this to make it to the nearest hundreth: 34.8(0)
That means that this is the correct order: 34.8, 36.43, 36.29 or Option B

5 0

3 years ago

Read 2 more answers

Which expression has the same value as 4x – 3y?

coldgirl [10]

Answer:

4x+(-3y)

Step-by-step explanation:

$\\ \rm\longmapsto 4x+(-3y)$

(-)(-)=(+)
(+)(-)=(-)
(-)(+)=(-)
(+)(+)=(+)

$\\ \rm\longmapsto 4x-3y$

5 0

3 years ago

Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is 13.4 years with a s

Aliun [14]

Answer : 0.0129

Step-by-step explanation:

Given : Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is $\mu=13.4$ years and standard deviation is $\sigma=1.7$ years.

Sample size : $n=40$

Let X be the random variable that represents the age of fleets.

We assume that the ages of the fleets of the 10 largest U.S. commercial passenger carriers are normally distributed.

For z-score,

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x=14

$z=\dfrac{14-13.4}{\dfrac{1.7}{\sqrt{40}}}\approx2.23$

By using the standard normal distribution table , the probability that the average age of these 40 airplanes is at least 14 years old will be :-

$P(x\geq 14)=P(z\geq2.23)\\\\=1-P(z$

Hence, the required probability = 0.0129

6 0

3 years ago

The population of Groverdale is

wolverine [178]

Answer: Yes alaina is correct

Step-by-step explanation: first look at the populations

553,000

535,841

Tittlehook has a 3 in the ten thousands place whilst groverdale has a 5 in the thousands place.

Hope this helps :)

6 0

3 years ago