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3 years ago

8

A video streaming service offers unlimited movies for $15 per month or $1.99 per movie. Write an inequality that represents when

the unlimited offer is better.

1 answer:

stira [4]3 years ago

3 0

If you watch let's say a movie a day all through July.(31 days) you'd pay if you round of $62 instead of $15 which is less than half the price.

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32% of =8 please. Thank you.

Oksi-84 [34.3K]

Answer:

2.56

Step-by-step explanation:

8* 32=256

256/100=2.56

6 0

3 years ago

PLS HELP 50 POINTS!!

Leno4ka [110]

Step-by-step explanation:

everything can be found in the picture above. Hope this helps

7 0

3 years ago

Read 2 more answers

The following equation is true for all values of x. Write the number that completes the equation. Show your work.

labwork [276]

5y+2(3y-1)=__y-(y+2)
5y+6y-2=__y-y-2
11y-2+2+y=__y
12y=___y
so 12

6 0

3 years ago

Which expression is equal to 2−√ 32 ?<br> 8i√2<br><br> −8√2<br><br> 8√2i<br><br> −8i√2

sdas [7]

$2 \sqrt{-32}=2 \sqrt{-1*16*2} =2 \sqrt{-1} * \sqrt{16} * \sqrt{2} =2i*4* \sqrt{2} =8i \sqrt{2}$

I hope that helps!

7 0

4 years ago

Read 2 more answers

The Institute of Management Accountants (IMA) conducted a survey of senior finance professionals to gauge members’ thoughts on g

DochEvi [55]

Answer:

(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

Step-by-step explanation:

Let <em>X</em> = number of senior professionals who thought that global warming is having a significant impact on the environment.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 100 and <em>p</em> = 0.65.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>p</em> if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10$

Thus, a Normal approximation to binomial can be applied.

So, $\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)$ .

(1)

Compute the value of $P(0.64$ as follows:

$P(0.64$

$=P(-0.20$

Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 90%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of <em>z</em> for P (Z < z) = 0.95 is

<em>z</em> = 1.65.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73$

Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 95%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of <em>z</em> for P (Z < z) = 0.975 is

<em>z</em> = 1.96.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75$

Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

7 0

4 years ago