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faltersainse [42]

2 years ago

13

Katherine bought a total of 14/16 pound of blue beads and silver beads at the store. The weight of the blue beads is 6/8 of the

total weight.what is the weight of the blue beads?

2 answers:

Stolb23 [73]2 years ago

7 0

Answer:

it D 12/16

step-by-step explanation:

when doing that kind of thing think if you can multiply the smaller denominator buy something to get the larger denominator so in this case you had to multiply the denominator by 2 to get 16. so after you're done with that you mutiply the numerator by how many you multiplied the smaller denominator to get the numerator of the new fraction so in this case you multiply 6 by 2 to get 12. so the end result would be 12/16

nadya68 [22]2 years ago

3 0

Answer:

Answer D

Step-by-step explanation:

Because you need to find a common denominator 6x2=12 8x2=16 so it 12/16

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Ben and Landon took turns driving on their recent 820 mile road trip. Ben averaged 60 miles per hour while landon averaged 56 mi

Furkat [3]

Landon drove for 5 hours.

<u>Step-by-step explanation</u>:

Given that,

Total distance of the road trip = 820 miles

Speed of Ben = 60 miles per hour

Speed of Landon = 56 miles per hour

<u><em>Step 1 </em></u><em>:</em>

Let 'x' be the number of miles Ben drove

Let 'y' be the number of miles Landon drove

<u><em>Step 2</em></u><em> :</em>

The trip took them 14 hours to complete 820 miles together.

x + y = 14

60x + 56y = 820

<u><em>Step 3 </em></u><em>:</em>

Solve the linear system x+ y= 14 and 60x+ 56y= 820

x + y = 14

x = 14 - y

<u><em>Step 4 </em></u><em>:</em>

Substitute x= 14-y in the second equation,

60(14-y) + 56y = 820

840 - 60y + 56y = 820

<u><em>Step 5</em></u><em> :</em>

Keeping y terms on one side and constants on other side,

60y - 56y = 840 - 820

4y = 20

y = 20/4

y = 5

Since 'y' is the number of miles Landon drove, the time Landon drove is 5 hours.

3 0

2 years ago

Nick is 30 years less than 3 times Rays age. If the sum of their ages is 74,how old are each of the men? Solve and write an equa

zaharov [31]

Nick = n
Ray = r

n=3r-30
n+r=74

3r-30+r=74
4r-30=74
4r=104
r=26
n=26*3-30
n=78-30
n=48
Nick is 48 years old and Ray is 26 years old
Hope this helps.

5 0

2 years ago

What is the sum of 3/2x and 7/4x

Advocard [28]

3/2x +7/4x = ...........

3/2x(4/4)+7/4x(2/2)

12/8x+14/8x = 26/8x

26/8x = 13/4x(3.25x)

4 0

2 years ago

Read 2 more answers

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability, using continuity correction, is $P(X < 12 - 0.5) = P(X < 11.5)$ , which is the pvalue of Z when X = 11.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0

2 years ago

As a promotion, the first 50 customers who entered a certain store at a mall were asked to choose from one of two discounts. The

qaws [65]

Answer:

A) Yes, because P (F∩S) = 0

Step-by-step explanation:

Hello!

50 customers of a store were asked to choose between two discounts:

Discount 1: 20% off all purchases for the day.

Discount 2: 10% off all purchases for the week.

28 choose discount 1

22 choose discount 2

F: the selected person choose discount 1.

S: the selected person choose discount 2.

Two events are mutually exclusive when the occurrence of one of them prevents the other from occurring in one repetition of the trial and the intersection between these two events is void with zero probability of happening.

In this case, since the customers were asked to choose one out of the two events, if the customer chooses the first one, then he couldn't have chosen the second one and vice-versa. Then the intersection between these two events has zero probability, symbolically:

P(F∩S)=0

I hope it helps!

8 0

2 years ago