Ask question

Ask question

All categories

1 year ago

6

The vet told Jake that his dog, Rocco, who weighed 55 pounds, needed to lose 10 pounds. Jake started walking Rocco every day and

changed the amount of food he was feeding him. Rocco lost half a pound the first week. Jake wants to determine Rocco’s weight in pounds, p, after w weeks if Rocco continues to lose weight based on his vet’s advice.
The equation of the scenario is what.

The values of p must be what.

2 answers:

ohaa [14]1 year ago

6 0

Answer:

(on Edge) 1. "The equation of the scenario is p = 55 - 0.5w"

2. "The values of p must be any whole number 45 to 55"

BTW i got number 1 from this user brainly.com/profile/Alyssamarie03-2838077

jarptica [38.1K]1 year ago

4 0

Answer:

.5xW

Step-by-step explanation:

Times .5 by W and by the amount of weeks he does it for is the outcome

You might be interested in

James purchased five bonds of face value of $1,000 that paid 5% annual interest rate. the total annual interest income of james

Trava [24]

Given:
5 bonds of face value of 1,000 that paid 5% annual interest rate.

5 bonds x 1,000 = 5,000
5,000 x 5% x 1 year = 250

The total annual interest income of James is 250. Each bond earns 50 per annum.

8 0

1 year ago

Read 2 more answers

If x varies jointly as y and z, and x = 8 when y = 4 and z = 9, find z when x = 16 and y = 6. 3

Ghella [55]

Answer:

Joint variation says that:

if $x \propto y$ and $x \propto z$

then the equation is in the form of:

$x = kyz$ , where, k is the constant of variation.

As per the statement:

If x varies jointly as y and z

then by definition we have;

$x=k(yz)$ ......[1]

Solve for k;

when x = 8 , y=4 and z=9

then

Substitute these in [1] we have;

$8=k(4 \cdot 9)$

⇒ $8 = 36k$

Divide both sides by 36 we have;

$\frac{8}{3}=k$

Simplify:

$k = \frac{2}{9}$

⇒ $x = \frac{2}{9}yz$

to find z when x = 16 and y = 6

Substitute these value we have;

$16 = \frac{2}{9} \cdot 6 \cdot z$

⇒ $16 = \frac{12}{9}z$

Multiply both sides by 9 we have;

$144 = 12z$

Divide both sides by 12 we have;

12 = z

or

z = 12

Therefore, the value of z is, 12

7 0

1 year ago

Read 2 more answers

Suppose that a person's birthday is a uniformly random choice from the 365 days of a year (leap years are ignored), and one pers

maxonik [38]

P ( A ∩ B ∩ C) = 1/365
P(A) = 1/365, P(B)= 1/365, P(C) = 365
If events A,B and C are independed then P (A ∩ B ∩ C) = P (A) P(B) P(C) must be true,
From the probabilities we have
1/365≠ 1/365 * 1/365 * 1/365
Thus, events A,B, C are not independent.

3 0

1 year ago

Helena needs 3.5 cups of flour per loaf of bread and 2.5 cups of flour per batch of muffins. She also needs 0.75 cup of sugar pe

Arlecino [84]

From the given data, we can generate two equations with two unknowns.

We let x = number of loaves of bread
y = number of batches of muffins

For the equation of the flour requirement:
17 = 3.5x + 2.5y

<span>For the equation of the sugar requirement:
</span>4.5 = 0.75x + 0.75y

We evaluate the solutions by manipulating one of the equations into terms of the other. We use the first equation.We write x in terms of y.

x = (4.5/0.75) - y

Substitute the third equation to the second equation.

17 = (3.5((4.5/0.75)-y)) + 2.5y

Evaluating y and x, we have,

y = 4 and x = 2

Thus, from the amounts she has in hand, she can make 4 loaves of bread and 2 batches of muffins.

4 0

1 year ago

Read 2 more answers

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

$\large t_{0.05}=2.365$

Now we can compute our confidence interval

$\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546$

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

6 0

2 years ago