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scZoUnD [109]
2 years ago
7

A hot air balloon descends to the ground. The function h(t) = 210 – 15t can be used to describe the altitude of the balloon as i

t approaches the ground. Which statement best describes the graph of the function that models the descent of the balloon?
A) The graph is discrete because there cannot be fractional values for time.
B) The graph is discrete because there cannot be negative values for altitude.
C) The graph is continuous because there can be fractional values for time.
D) The graph is continuous because there can be negative values for altitude.
Mathematics
2 answers:
e-lub [12.9K]2 years ago
8 0

Answer:

C is the answer

Step-by-step explanation:

Olin [163]2 years ago
4 0
It is D if its correct i will explain why
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You and a classmate have a bug collection for science class. You find 5 out of every 9 bugs in the collection. You find 4 more bug
FrozenT [24]

Answer:

The total number of bugs in the collection is 36

Step-by-step explanation:

Let

x -----> the number of bugs that you find out

y ----> the number of bugs that the classmate find out

we know that

\frac{x}{y+x} =\frac{5}{9}  -----> equation A

x=y+4 ----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

The solution is the point (20,16)

The total number of bugs in the collection is

(20+16)=36\ bugs

5 0
2 years ago
An insurance company collected data from a class of high school sophomores on whether or not they have a cell phone and whether
NeX [460]

Answer:

C. Those who have a car tend to have a cell phone.

Step-by-step explanation:

The relative frequency for a data set n is calculated by dividing each frequency x_i by n.

We have a table that relates the use of cars and cell phones.

First, we take from the table the population that has a car.

n = 18

Then, of the students who have a car, 12 have a cell phone and 6 do not have a cell phone.

Then we calculate the relative frequencies f_{x_i} for those who have a cell phone and those who do not.

______________________________________________

Relative frequency   Students with car  n = 18.

-------------------------------------------------- --------------------------------------

C<em>ell phone     No cell phone      Total n</em>

 12/18                    6/18                     18

_______________________________________________

_______________________________________________

Relative frequency   Students with car  n = 18.

-------------------------------------------------- --------------------------------------

C<em>ell phone     No cell phone      Total n</em>

  0.666              0.333                   18

________________________________________________

Most students who have a car also have a cell phone.

Now we calculate the relative frequency for students who do not have a car.

_______________________________________________

Relative frequency  Students without a car  n = 7

-------------------------------------------------- --------------------------------------

<em>Cell phone       No cell phone             Total  n</em>

 2/7                     5/7                          7

_______________________________________________  

_______________________________________________

Relative frequency  Students without a car  n = 7

-------------------------------------------------- --------------------------------------

<em>Cell phone</em>      <em>  No</em> <em>cell phone           Total  n</em>

 0.286                 0.714                              7

_______________________________________________

Most students who do not have a car do not have a cell phone either.

<em>Then these data suggest that. Those who have a car tend to have a cell phone.</em>

Option C

6 0
2 years ago
The bookstore at State University purchases from a vendor sweatshirts emblazoned with the school name and logo. The vendor sells
svlad2 [7]

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

7 0
2 years ago
Andy, who loves to cook, makes apple cobbler for his family. The recipe (serves 6) calls for 434 pounds of apples, 514 cups of f
Tamiku [17]

This question was not written properly

Complete question;

Andy, who loves to cook, makes apple cobbler for his family. The recipe (serves 6) calls for 4 3/4 pounds of apples, 5 1/4 cups of flour, 1/6 cup of margarine, 2 7/8 cups of sugar, and 2 teaspoons of cinnamon. Since guests are coming, Andy wants to make a cobbler that will serve 15 (or increase the recipe 2 1/2 times). How much of each ingredient should Andy use?

Answer:

To serve 15, number of ingredients Andy should use =

Apples = 11 7/8 pounds

Flour = 13 1/8 cups

Margarine = 5/12 cups

Sugar = 7 3/16 cups

Cinnamon = 5 teaspoons.

Step-by-step explanation:

The recipe (serves 6) calls for 4 3/4 pounds of apples, 5 1/4 cups of flour, 1/6 cup of margarine, 2 7/8 cups of sugar, and 2 teaspoons of cinnamon. Since guests are coming, Andy wants to make a cobbler that will serve 15 (or increase the recipe 2 1/2 times). How much of each ingredient should Andy use?

For Apples

Serving for 6 = 4 3/4 pounds of apples

Serving for 15 =

= 15 × 4 3/4 pounds/6

= 11 7/8 pounds of apples

For flour

6 servings = 5 1/4 cups of flour,

15 servings =

15 × 5 1/4 cups/6

= 13 1/8 cups of flour

For Margarine

6 servings = 1/6 cups of margarine

15 servings =

= 15 × 1/6 cups/6

= 5/12 cups of margarine

For Sugar

6 servings = 2 7/8 cups of sugar

15 servings =

15 × 2 7/8 cups/6

= 7 3/16 cups of sugar

For Cinnamon

6 servings = 2 teaspoons of cinnamon

15 servings =

15 × 2/6

= 5 teaspoons

Therefore, to serve 15, number of ingredients Andy should use =

Apples = 11 7/8 pounds

Flour = 13 1/8 cups

Margarine = 5/12 cups

Sugar = 7 3/16 cups

Cinnamon = 5 teaspoons.

5 0
2 years ago
The weight of corn chips dispensed into a 14-ounce bag by the dispensing machine has been identified as possessing a normal dist
mamaluj [8]
The probability that a normally distributed dataset with a mean, μ, and statndard deviation, σ, exceeds a value x, is given by

P(X\ \textgreater \ x)=1-P(X\ \textless \ x)=1-P\left(z\ \textless \  \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \right)

Given that t<span>he weight of corn chips dispensed into a 14-ounce bag by the dispensing machine is a normal distribution with a mean of 14.5 ounces and a standard deviation of 0.2 ounce.

</span>If <span>100 bags of chips are randomly selected the probability that the mean weight of these 100 bags exceeds 14.6 ounces is given by

P(X\ \textgreater \ 14.6)=1-P\left(z\ \textless \  \frac{14.6-14.5}{\frac{0.2}{\sqrt{100}}} \right) \\  \\ =1-P\left(z\ \textless \  \frac{0.1}{\frac{0.2}{10}} \right)=1-P\left(z\ \textless \  \frac{0.1}{0.02} \right) \\  \\ =1-P(z\ \textless \ 5)=1-1=0

Therefore, the probability that </span><span>the mean weight of these 100 bags exceeds 14.6 ounces is</span> 0.

7 0
2 years ago
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