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ale4655 [162]
3 years ago
13

Please help so I can make my family proud

Mathematics
1 answer:
Levart [38]3 years ago
8 0

Answer:

A = 83.5m^2

Step-by-step explanation:

9 * 19 = 171m for the complete rectangle

9 - 4 = 5

5 * 19 = 95

95 / 2 = <u>47.5m</u> for the triangles

9 * 4 = <u>36m</u>

36 + 47.5 = <u>83.5m</u>

<em><u>Plz mark as brainliest if correct! Have a nice day!!! </u></em>

<em><u>-Lil G</u></em>

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Alexis put $2000 in a savings account. After 4 years, she had $2543 in the account. What rate of interest did she earn? Use the
Flauer [41]
This is substitution, it's very simple once you get the hang of it! 

okay, what you want to do is plug in the numbers you already have, to get the number you don't have(r).

First, you plug in the numbers.

1.  P= $2000  
    A=$2543
    t= 4(years)
    r=? 

2543=2000^r(4) 

Secondly, you want to solve the equation.


5 0
3 years ago
Read 2 more answers
Determine what type of model best fits the given situation:
lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = \frac{20}{100} \times P = 0.2P

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = \frac{20}{100} \times 0.8P = (0.2)(0.8P)

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

=P(0.8)^{2}-------------------------(2)

Similarly after 3 years, this value P(0.8)^{2} will again be decreased by 20 %

so 20% of P(0.8)^{2}  \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}

So after 3 years value is decreased by (0.2)P(0.8)^{2}

so value after 3 years will be = P(0.8)^{2}   - (0.2)P(0.8)^{2}

taking P(0.8)^{2} common out we get P(0.8)^{2}(1-0.2)

P(0.8)^{2}(0.8)

P(0.8)^{3}-----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or P(0.8)^{1}

value after 2 years is P(0.8)^{2}

value after 3 years is P(0.8)^{3}

so value after x years will be P(0.8)^{x} ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

y = P(0.8)^{x} where y is the value after x years

so thats the final answer

3 0
3 years ago
The time needed to complete a final examination in a particular college course is normally distributed with a mean of 77 minutes
Komok [63]

Answer:

a. The probability of completing the exam in one hour or less is 0.0783

b. The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. The number of students will be unable to complete the exam in the allotted time is 8

Step-by-step explanation:

a. According to the given we have the following:

The time for completing the final exam in a particular college is distributed normally with mean (μ) is 77 minutes and standard deviation (σ) is 12 minutes

Hence, For X = 60, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=60−77 /12

Z=−1.4167

Using the standard normal table, the probability P(Z≤−1.4167) is approximately 0.0783.

P(Z≤−1.4167)=0.0783

Therefore, The probability of completing the exam in one hour or less is 0.0783.

b. In this case For X = 75, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=75−77 /12

Z=−0.1667

Using the standard normal table, the probability P(Z≤−0.1667) is approximately 0.4338.

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is obtained as follows:

P(60<X<75)=P(Z≤−0.1667)−P(Z≤−1.4167)

=0.4338−0.0783

=0.3555

​

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. In order to compute  how many students you expect will be unable to complete the exam in the allotted time we have to first compute the Z−score of the critical value (X=90) as follows:

Z=  X−μ /σ

Z=90−77 /12

Z​=1.0833

UsING the standard normal table, the probability P(Z≤1.0833) is approximately 0.8599.

Therefore P(Z>1.0833)=1−P(Z≤1.0833)

=1−0.8599

=0.1401

​

Therefore, The number of students will be unable to complete the exam in the allotted time is= 60×0.1401=8.406

The number of students will be unable to complete the exam in the allotted time is 8

6 0
3 years ago
Do you think x and x by the power of 2 are like terms? Explain using words, diagrams, tables etc.
natali 33 [55]

Answer:

No

Step-by-step explanation:

No, because x + x^2 cannot be grouped together.

8 0
4 years ago
1/8 - 1/7 please :DDDDDDDDD
ira [324]

Answer:

-1/56

Step-by-step explanation:

6 0
3 years ago
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