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Maru [420]
3 years ago
8

Leo had a $30 gift card to use at a coffee shop. He used the entire balance to buy regular-sized coffees, which cost $2, and lar

ge coffees, which cost $3. If he bought 13 coffees in total, how many large coffees did he buy?
Select all that apply.
A. 2x+3y=30
B. 2x+3y=13
C. x+y=13
D. x+y = 30
Mathematics
2 answers:
bija089 [108]3 years ago
5 0

Answer:

B

Step-by-step explanation:

djyliett [7]3 years ago
4 0

Answer  4

If he bought 4 large coffees that means he spent $12 on them. that leaves $18. Which means he can buy 9 regular sized coffees. All together thats 13 coffees.

Step-by-step explanation:

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Answer:

wvx=57

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26

8 0
3 years ago
Can someone help me find the equivalent expressions to the picture below? I’m having trouble
miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

           =2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}

           =2^{\frac{2}{9}}\times 3^{-\frac{2}{3} } [Option 2]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0
3 years ago
Graph + > -1.
andrew-mc [135]

You have the right idea but the endpoint is at the wrong location. Instead, the green dot should go at -1. The shading is to the right. We use a closed filled in circle here (instead of an open hole) to tell the reader "include the endpoint". So -1 is part of the solution set.

In short, the graph consists of a closed filled in circle at -1 with shading to the right. This visually describes all values that are larger than -1, or equal to -1.

6 0
3 years ago
Read 2 more answers
A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.57 and the standard deviation is $20.28.
son4ous [18]

Answer:

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.01 = 0.99, so z = 2.325

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.325*\frac{20.28}{\sqrt{20}} = 12.17

The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.

The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

3 0
3 years ago
water is piped from a dam to a town 24 km away using pipes each 8m long. How many pipes will be needed?
sineoko [7]
A = cross sectional area of the pipe. When the pressure is expressed in terms of the equivalent height of a column of that fluid, as is common with water, the friction loss is expressed as S, the "head loss" per length of pipe, a dimensionless quantity also known as the hydraulic slope.
3 0
3 years ago
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