Complete the equation for this model.

I really need help would please like help thanks

ASHA 777 [7]

Hey there!

A reciprocal might be an unfamiliar vocab word, but the concept is really quite simple.

It's just a flipped fraction! Multiplying the flipped fraction by it's reciprocal gets you one, and that's how you check your answer.

So we have:

Reciprocal of 9/10 is 10/9, because 9/10 times 10/9 = 90/90 = 1

Reciprocal of 1/11 is 11/1, because 1/11 times 11/1 = 11/11 = 1

Reciprocal of 10(remember, it's 10/1) is 1/10, because 10/1 times 1/10 = 10/10 = 1

Hope this helps!

4 0

2 years ago

If an orthocenter lies inside of a triangle, then the triangle must be

Black_prince [1.1K]

Answer:

<h2><u><em>acute</em></u></h2>

Step-by-step explanation:

Remember :the orthocenter of a triangle is the point of concurrency of the 3 altitudes.

$\large \text{If an orthocenter lies inside of a triangle, then the triangle must be}\ :\\\\\large isosceles. \\\\\large obtuse. \\\\\large right. \\\\\large acute \checkmark$

8 0

2 years ago

Devon and his friends bought strawberry wafers for $3 per packet and chocolate wafers for $1 per packet at a carnival. They spen

Irina-Kira [14]

Hi there!

PART A:
The system of equations we would use would be:
x + y = 22 (amount of items)
3x + 1y = 30 (cost)

Variables:
x = the amount of strawberry wafers bought at the price of $3
y = the amount of chocolate wafers bought at the price of $1

PART B:
To solve, we'll use substitution because we can easily isolate a variable using the first equation.

Work:
x + y = 22 (first equation)
y = 22 - x (isolating a variable)
3x + 1y = 30 (second equation)
3x + (22 - x) = 30 (substituting into the second equation)
2x + 22 = 30 (simplifying)
2x = 8 (subtracting)
x = 4 strawberry wafers
4 + y = 22 (substituting x into the first equation to solve for y)
y = 18 chocolate wafers

ANSWER:
They bought 4 strawberry wafers and 18 chocolate wafers.

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

8 0

3 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

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

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

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

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

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

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

3 years ago

Peter cycles for 1/4 hours at a speed of 20 km/h

yKpoI14uk [10]

Answer:

His average speed is

12 km/hr

.

Step-by-step explanation:

Remember the triangle of Speed, Distance and Time. If you remember it, you'll ace these kinda questions.

I had trouble with these formulas but the triangle helped me a LOT! Anyways, let's get back to the question. The formula for average speed is pretty much the same as the formula for just speed. The average speed formula is

Average speed

=

Total Distance

Total Time

So......

48 km

4 hr

=

12 km/hr

My source

I hope this explanation helps you!

5 0

2 years ago