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Lemur [1.5K]
3 years ago
10

In the past month, Eric rented 1 video game and 6 DVDs. The rental price for the video game was $2.20. The rental price for each

DVD was 3.70 What is the total amount that Eric spent on video game and DVD rentals in the past month?​
Mathematics
2 answers:
snow_lady [41]3 years ago
3 0
The total amount of money Eric spent is $24.4 dollars! Hope this helps!
rusak2 [61]3 years ago
3 0
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What is inequality x - 4 ‹ 0
lesya [120]

You simply add 4 to 0.

x-4<0

x<4


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Does anybody know this plz help.
MaRussiya [10]
So for this it's asking to multiply 10,000 glass beads by the length of 0.4 cm.

So to make it easier let's just make it a fraction.

This will make it 4/10 (because 0.4 is just 4 tenths of a whole number)

So now we can upscale the sizes to the proper ratio. This ratio being the thousands.

4000/10000

Meaning the answer is 4000cm in length.
3 0
3 years ago
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Rewrite the equation in vertex form. Name the vertex and y-intercept. y=3/5x^2+30x+382
Serga [27]
<span>(y-7) = 3/5 (x+25)^2 so the y intercept is -7 and the vertex is 35

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7 0
3 years ago
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John borrowed $150,500 to start up his business. He borrowed the money at a 7.75% simple
zalisa [80]

Answer:

John have to pay back $23,327.50 at the end of two years

Step-by-step explanation:

Simple Interest (I) = (PRT) ÷ 100

where P = Principal, R = Rate, T= Time

I = ($150500 × 7.75 × 2) ÷ 100

I = ($2332750) ÷ 100

I = $23,327.50

6 0
3 years ago
A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard
bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44

In point a)  

\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

  = \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2

value= (1- \frac{1}{k^2}) \times 100 \% =75\%

In point b)

\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

In point c)

\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

In point d)

using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

8 0
3 years ago
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