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siniylev [52]
3 years ago
5

John's goal is to have more then -7 dollars in his bank account by the end of the month.The variable d is the number of dollars

in John's bank account at the end of the month.
Write a inequality in terms of d that is true only if John meets his monthly goal.

HELP!!!! IF U ANSWER THE QUESTION ILL GIVE U BRAINLYIEST
Mathematics
2 answers:
Alona [7]3 years ago
6 0

Answer:

i need more info

Step-by-step explanation:

fredd [130]3 years ago
3 0
I think it would be
D ≥ -7 but I’m not too sure
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What is the sum of two numbers is 45 and theire difference is 5
adelina 88 [10]

Answer:

let the two numbers be x and y

x+y=45(equation 1)

x-y=5(equation 2)

from equation (1) y=45-x(equation 3)

substitute 45-x for y in equation 2

x-(45-x)=5

x-45+x=5

x+x-45=5

2x-45=5

2x=45+5

2x=50

x=50/2

x=25

6 0
3 years ago
First chance you get the best marks ​
garri49 [273]

Answer:

<h2><em><u>First box and last box.</u></em></h2>

Step-by-step explanation:

<h2><em><u>It is the first box.</u></em></h2>

Distribute 7 to the first number

7 * -3/4 = -21/4

-21/4 = 5  -1/4

Distribute 7 to the second number

7 * -3

= -21

Put the numbers together

<h2><em><u>5  -1/4x  -21 is the answer.</u></em></h2>

<h2><em><u>It is also the last box.</u></em></h2>

They separated the parenthesis.

So it is still 7 * -3/4

and

7*-3.

Hope this helps,

Kavitha

7 0
4 years ago
The mean height of an adult giraffe is 17 feet. Suppose that the distribution is normally distributed with standard deviation 0.
BARSIC [14]

Answer:

Median =17

Step-by-step explanation:

Given that the mean height of an adult giraffe is 17 feet. Suppose that the distribution is normally distributed with standard deviation 0.9 feet.

Let X be the height of a randomly selected adult giraffe.

X is a random variable.  

a) X is N(17, 0.9)

b) Median giraffe height = 17 ft since in normal distribution mean = median

c) When x = 18.5, Z = \frac{18.5-17}{0.9} =0.1667

d) P(X

e) P(16.9

f) 75th percentile for z is 0.675

x=17.9+0.9(0.675)\\= 18.5075

5 0
3 years ago
12. The probability a basketball player making a free-throw
Rasek [7]

Answer:

The odds against making a free throw are: 0.1905 or 0.16:0.84

Step-by-step explanation:

The odds can be categorized into two categories

  • Odds-against
  • Odds-in favour

The odds against is the ratio of probability of failure to probability of success.

Given

Probability of making a free throw = P = 0.84

Probability of not making a free throw = Q = 1-P = 1-0.84 = 0.16

The formula for odds-against is:

Odds-against = \frac{0.16}{0.84}\\= 0.19047..\\=0.1905

Hence,

The odds against making a free throw are: 0.1905 or 0.16:0.84

5 0
3 years ago
What are the vertices of triangle PQR?
deff fn [24]
I believe it’s D I’m not 100 percent sure tho
4 0
3 years ago
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