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3 years ago

5

At a deli, 56 sandwiches were sold during lunchtime. Twenty-five percent of the sandwiches sold were tuna salad sandwiches. How

many of the sandwiches sold were not tuna salad?

1 answer:

denpristay [2]3 years ago

8 0

Answer - 42 Sandwiches

Explanation -
56 is what you start out with. you will need to find 25% of those 56 to see the remaining. So 25% of 56 is 14 because .25 x 56 is 14. You then subtract 56 from the 14 Tuna sandwiches and you’re left with 42.

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Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

3 years ago

URGENT URGENT helppps plz

Murljashka [212]

Answer:

1) Bigger sector = 168.8pi

1) Smaller sector = 56.3pi

2) Bigger sector = 529.9

3) Smaller sector = 176.6

Step-by-step explanation:

1) Bigger sector = $\frac{270}{360}$ ×pi×15×15

= 168.75pi

1) Smaller sector = $\frac{360-270}{360}$ ×pi×15×15

= $\frac{90}{360}$ ×pi×15×15

= 56.25pi

2) Bigger sector = $\frac{270}{360}$ ×3.14×15×15

= 529.875

2) Smaller sector = $\frac{360-270}{360}$ ×3.14×15×15

= $\frac{90}{360}$ ×3.14×15×15

= 176.625

6 0

3 years ago

Use the data in the table below to create a scatter plot. Based on your scatter plot, does the data

Rashid [163]

Answer:

C

Step-by-step explanation:

There is a positive correlation, as hours spent gets bigger the test score gets bigger with some outliers

8 0

2 years ago

At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is

Sphinxa [80]

Answer:

34.87% probability that all 5 have a wireless device

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

81% of students own a wireless device.

This means that $p = 0.81$

If 5 students are selected at random, what is the probability that all 5 have a wireless device?

This is P(X = 5) when n = 5. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.81)^{5}.(0.19)^{0} = 0.3487$

34.87% probability that all 5 have a wireless device

5 0

3 years ago

a researcher sets out to measure drug use on US college campuses by asking a representative sample of undergraduates whether the

Readme [11.4K]

Based on the information given, it can be noted that the problem with the research is the problem of validity.

<h3>Problems with validity.</h3>

It should be noted that in a research, a problem with validity takes place when there's a lack of reliability of the independent variable.

Also, it is a result of lack of representativeness of the independent variable and a lack of impact of the independent variable.

In this case, the researcher sets out to measure drug use on US college campuses by asking a representative sample of undergraduates whether they are currently receiving federal grants or loans. It should be noted that the question asked doesn't relate to the research.

In conclusion, there's a problem with validity.

Learn more about researches on:

brainly.com/question/25257437

5 0

2 years ago