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Vladimir [108]
3 years ago
13

How many solutions does this system of equations have?

Mathematics
2 answers:
Andrew [12]3 years ago
8 0

I wanna say and go with D but im not quiet sure


Studentka2010 [4]3 years ago
6 0
C because y=5x-4 has only one solution to it compared to 5x+2 that has 2 or more
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I need help please! I've been stuck on this for a week and its really stressful lol
worty [1.4K]

Answer:

y=0.2*4^{x}

Step-by-step explanation:

Notice when x increases 1, y is 4 times the previous one, so

the function is like y=C*4^{x}

To determine the constant C, put any pair of (x, y)

Use x = 0, y = 0.2, so

0.2 = C*4^{0} = C * 1 = C

then y=0.2*4^{x}

5 0
2 years ago
Read 2 more answers
For the years from 2002 and projected to 2024, the national health care expenditures H, in billions of dollars, can be modeled b
dmitriy555 [2]

Answer:

2019.

Step-by-step explanation:

We have been given that for the years from 2002 and projected to 2024, the national health care expenditures H, in billions of dollars, can be modeled by H = 1,500e^{0.053t} where t is the number of years past 2000.

To find the year in which national health care expenditures expected to reach $4.0 trillion (that is, $4,000 billion), we will substitute H=4,000 in our given formula and solve for t as:

4,000= 1,500e^{0.053t}

\frac{4,000}{1,500}=\frac{ 1,500e^{0.053t}}{1,500}

\frac{8}{3}=e^{0.053t}

e^{0.053t}=\frac{8}{3}

Take natural log of both sides:

\text{ln}(e^{0.053t})=\text{ln}(\frac{8}{3})

0.053t\cdot \text{ln}(e)=\text{ln}(\frac{8}{3})

0.053t\cdot (1)=0.9808292530117262

\frac{0.053t}{0.053}=\frac{0.9808292530117262}{0.053}

t=18.506212320

So in the 18.5 years after 2000 the expenditure will reach 4 trillion.

2000+18.5=2018.5

Therefore, in year 2019 national health care expenditures are expected to reach $4.0 trillion.

7 0
3 years ago
A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their pur
Harrizon [31]

Answer:

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

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 - 0.8}{2} = 0.1

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

That is z with a pvalue of 1 - 0.1 = 0.9, so Z = 1.28.

Now, find the margin of error 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 = 1.28\frac{1.5}{\sqrt{305}} = 0.1

The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5

The upper end of the interval is the sample mean added to M. So it is 7.6 + 0.1 = 7.7

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

8 0
3 years ago
Which inverse operation would be used to verify the following equation?
sveticcg [70]
Another one is 34 times 3
8 0
3 years ago
Read 2 more answers
Help------------------------------------
Liula [17]

Answer:

7)141

8)65

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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