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3 years ago

Write an equation for each problem. Then solve.

Mathematics

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

<em>The price is the same at both stores for 2 prints.</em>

Step-by-step explanation:

<u>Equations</u>

Let's set the variable

x = number of photo prints

Company Photo Plus charges $2 for each print and $6 for a processing fee, thus the total charges are:

PP = 6 + 2x

Company Picture Time charges $3 for each print and $4 for a processing fee, thus it charges a total of:

PT = 4 + 3x

It's required to find the number of prints that make both stores charge the same. Equating both functions:

6 + 2x = 4 + 3x

Subtracting 2x and 4:

x = 2

The price is the same at both stores for 2 prints.

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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Mike is 3 years old.joe is 6 times as old as mike. whict equation shows how to find Joe's age? 6÷3=2. 9-3=6. 3×6=18. 3×6=9

grin007 [14]

Answer: 18

Step-by-step explanation:

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4 0

3 years ago

A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e

Andre45 [30]

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$ , and $\frac{\alpha}{2}=0.025$ . Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

8 0

3 years ago

Is 2/3 greater then1/5

Greeley [361]

Yes, it is. To find this out, you need to make the denominators the same. They can both multiply into 15, so we change the denominators to 15. Whatever we do to the bottom, we also have to do to the top.

2/3 = 10/15

1/5 = 3/15

We can then see that 10 is more than 3 :)

8 0

4 years ago