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Tatiana [17]
1 year ago
9

Brett has consumed 1,400 calories so far today. He has also burned off 400 calories at the gym. He would like to keep his daily

calorie total to 2,000 calories per day. How many calories does he have left to consume for the day? Is 1,200 a viable solution to this problem?
I NEEED HELPP PLZZ :(
A Yes; 1,200 is less than 1,400.
B Yes; 1,200 is less than 2,000.
C No; 1,200 is more than the 400 he burned off at the gym.
D No; 1,200 will cause him to exceed 2,000.
Mathematics
1 answer:
Irina-Kira [14]1 year ago
3 0

Answer

D

Step-by-step explanation:

D is correct because 1,400-400= 1,000

so he can only consume 1,000 more calories so 1,200 will cause him to go over 2,000

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Identify the type of function graphed to the right. linear exponential other For an increase of 1 in the x-value, what is the in
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Melissa wants to make a table representing the area of her vegetable garden for a variety of different lengths of the tomato pat
Ksenya-84 [330]

Answer:

\begin{array}{cc}Length \ of \ Tomato \ Patch \ (in \ feet)&Area \ of \ Vegetable \ Garden \ (in  \ square \ feet)\\\ [6.25]&338\\6.5&[242.625]\\  \ [6.75]&147.25\\7&[51.875]\end{array}

Step-by-step explanation:

The date from the given table, is expressed as follows;

Area of Vegetable Garden (in square feet); 338, ___, 147.25, ___

Length of Tomato Patch (in feet); ___, 6.5, ___, 7

The length of the tomato patch increases with decrease in the area of the garden

Given that the length of the tomato patch is the independent variable, we can have;

Length of Tomato Patch (in feet); 6.25, 6.5, 6.75, 7

Therefore, for an increase in the length of the tomato patch from 6.25 to 6.75, (a change of Δl = 0.5) the area of the vegetable garden decreased from 338 to 147.25 which is a decrease of ΔA = 190.75

Therefore, the width of the tomato patch, w = ΔA/Δl = 190.75/0.5 = 381.5

The width of the tomato patch, w = 381.5 ft.

The relationship between the total area, TA, the area of the vegetable garden, <em>A</em>, and the length of the tomato patch, <em>l</em>, is therefore, given as follows;

A = TA - 381.5·l

338 = TA - 381.5×6.25

TA = 338 + 381.5×6.25 = 2,722.375

Therefore, when l = 6.5, we get

A = 2,722.375 - 381.5×6.5 = 242.625

When l = 7, we get

A = 2,722.375 - 381.5×7 = 51.875

Therefore, we get;

\begin{array}{cc}Length \ of \ Tomato \ Patch \ (in \ feet)&Area \ of \ Vegetable \ Garden \ (in  \ square \ feet)\\\ [6.25]&338\\6.5&[242.625]\\  \ [6.75]&147.25\\7&[51.875]\end{array}

5 0
1 year ago
In 1998 the average income for middle class families in the US was $37,100 with a population standard deviation of $6362. We wan
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Answer:

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

The best answer would be:

A. Z test

Step-by-step explanation:

Data given and notation  

\bar X=36670 represent the mean average

\sigma=6362 represent the population standard deviation for the sample  

n=1225 sample size  

\mu_o =37100 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal or not to 37100, the system of hypothesis would be:  

Null hypothesis:\mu =37100  

Alternative hypothesis:\mu \neq 37100  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

The best answer would be:

A. Z test

5 0
2 years ago
A survey found that 73% of adults have a landline at their residence (event A); 83% have a cell phone (event B). It is known tha
4vir4ik [10]

Answer:

3. What is the probability that an adult selected at random has both a landline and a cell phone?

A. 0.58

4. Given an adult has a cell phone, what is the probability he does not have a landline?

C. 0.3012

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that an adult has a landline at his residence.

B is the probability that an adult has a cell phone.

C is the probability that a mean is neither of those.

We have that:

A = a + (A \cap B)

In which a is the probability that an adult has a landline but not a cell phone and A \cap B is the probability that an adult has both of these things.

By the same logic, we have that:

B = b + (A \cap B)

The sum of all the subsets is 1:

a + b + (A \cap B) + C = 1

2% of adults have neither a cell phone nor a landline.

This means that C = 0.02.

73% of adults have a landline at their residence (event A); 83% have a cell phone (event B)

So A = 0.73, B = 0.83.

What is the probability that an adult selected at random has both a landline and a cell phone?

This is A \cap B.

We have that A = 0.73. So

A = a + (A \cap B)

a = 0.73 - (A \cap B)

By the same logic, we have that:

b = 0.83 - (A \cap B).

So

a + b + (A \cap B) + C = 1

0.73 - (A \cap B) + 0.83 - (A \cap B) + (A \cap B) + 0.02 = 1

(A \cap B) = 0.75 + 0.83 - 1 = 0.58

So the answer for question 3 is A.

4. Given an adult has a cell phone, what is the probability he does not have a landline?

83% of the adults have a cellphone.

We have that

b = B - (A \cap B) = 0.83 - 0.58 = 0.25

25% of those do not have a landline.

So P = \frac{0.25}{0.83} = 0.3012

The answer for question 4 is C.

4 0
2 years ago
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