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Bumek [7]
2 years ago
8

QUESTION ON PIC THX!

Mathematics
2 answers:
Nana76 [90]2 years ago
7 0

Answer: So you basically subtract 67 by 20% of it so that is

67-20%

Which is 53.6.

So the answer is $53.60

mojhsa [17]2 years ago
3 0

Answer:

The answer is $53.6

Step-by-step explanation:

Here,

67 × 20 ÷ 100 = 13.4

Now,

$67 – $13.4 = $53.6

Thus, The current price of the item is $53.6

<u>-TheUnknownScientist 72</u>

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If F(x) =3x-2/6 which of the following is the inverse of f(x)
marissa [1.9K]

Answer:

f^{-1}(x)=y/3-1/9

Step-by-step explanation:

f(x)=3x-2/6

f^{-1}(x)=x/3-1/9



if f(x)=(3x-2)/6

answer:f^{-1}(x)=(6x+2)/3

4 0
3 years ago
Bob withdrew money from an ATM, put the receipt in an envelope, and filled in the information in his check register.
lora16 [44]

Answer:It is important to record each transaction so Bob knows how much money is in his account. That prevents being overdrawn. Recording debits and deposits helps with budgeting. It is good to keep receipts in case of any bank errors.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
The heights of young women aged 20 to 29 follow approximately the N(64, 2.7) distribution. Young men the same age have heights d
murzikaleks [220]

Answer: 10.703%

Step-by-step explanation:

Let minimum height of the tallest 25% of young women be M.

Let Q be the random variable which denotes the height of young women.

Therefore, Q – N(64,2.70)

Now, P(Q˂M) = 1-0.25

i.e. P[(Q-64)/2.7 ˂ (M-64)/2.7] = 0.75

I.e. ф-1 [(M-64)/2.7] = 0.75 i.e. (M-64)/2.7 = ф-1 (0.75) = 0.675 i.e. M = 65.8198 inches

Let R be the random variable denoting the height of young men

Therefore, R – N (69.3, 2.8)

i.e. (R-69.3)/2.8 – N(0,1)

therefore the probability required = P(R ˂65.8198) = P[(R-69.3)/2.8 ˂ (65.8198 – 69.3)/2.8]

this gives P[(R-69.3)/2.8 ˂] = ф (-1.2429) = 0.107033

From this, the percentage of young men shorter than the shortest amongst the tallest 25% of young women is 10.703%

4 0
3 years ago
Please help if you can pic attatched
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

Subtract row 3 from row 1:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]

Subtract row 3 from 7 times row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]

Divide row 3 by 17:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 2 of row 3 from row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 6 of row 2 from row 1:

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]

C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]

3 0
3 years ago
On an average day, a garden snail can travel about 0.05 mile. The snail travels 0.2 times as far as the average distance on day
daser333 [38]

Answer: 0.04 miles

 

SOLUTION

 

Given,

Average daily distance of a garden = 0.05 mile.

 

Distance traveled on day 1 = 0.2 times as far as the average distance

= 0.2 x 0.05 miles

= 0.01 miles

 

Distance traveled on day 2 = 0.6 times as far as the average distance

= 0.6 x 0.05 miles

= 0.03 miles

 

Total distance traveled in two days

= Distance traveled on day 1 + Distance traveled on day 2

= 0.01 miles + 0.03 miles

= 0.04 miles

6 0
3 years ago
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