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3 years ago

8. Jordyn bought a $50 cell phone case that was discounted 15%. Taxes were 5%. What was her total

Mathematics

1 answer:

shepuryov [24]3 years ago

3 0

$50 x .15% = $7.50

$50 - $7.50 = $42.50

$50 x .05% = $2.50

$42.50 + $2.50 = $45

Answer : $45

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2- 1/2 divided by 1 - 1/4

wlad13 [49]

Answer: 1.25

Step-by-step explanation:

3 0

3 years ago

Find the greatest common factor of the expressions.<br> 5x^7, 30x

kogti [31]

To find the GCF (greatest common factor) we must find the largest thing that can go into both of them.

We cannot divide by any exponents because only one has them, however we can divide by 5x.

5x^7, 30x

x^7, 6

So, the GCF is 5x

Hope this helps,

Jeron

P.S

If this helps, consider marking brainliest

edit:

Thank you :D

5 0

3 years ago

Read 2 more answers

4(6 – 5x)+8(4 – 5x)<br> Please help

strojnjashka [21]

The answer is -60x + 56

4(6 - 5x) + 8(4 - 5x)

24 - 20x + 32- 40x

56 - 60x or you can write it as -60x + 56

3 0

3 years ago

Use the definition of continuity to determine whether f is continuous at a. f(x) = 5x+5 a = -5 Question

Andrej [43]

ANSWER

$lim_{x \to - 5}(f(x)) = f( - 5)$

EXPLANATION

If f(x) is continuous at

$x = a$

Then,

$lim_{x \to a}(f(x)) = f(a)$

The given function is

$f(x) = 5x + 5$

$f( - 5) = 5( - 5) + 5$

$f( - 5) = - 25 + 5 = - 20$

$lim_{x \to - 5}(f(x)) = 5( - 5) + 5$

$lim_{x \to - 5}(f(x)) = - 20$

Since,

$lim_{x \to - 5}(f(x)) = f( - 5)$

The function is continuous at

$x = - 5$

4 0

3 years ago

Read 2 more answers

Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500

Alborosie

Answer:

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 500, \pi = \frac{421}{500} = 0.842$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 - 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.81$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 + 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.874$

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

6 0

3 years ago