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marissa [1.9K]
3 years ago
12

Find the value of x

Mathematics
1 answer:
saw5 [17]3 years ago
8 0

Answer:

x=50

Step-by-step explanation:

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in a walk for charity, you walk at a rate of 100 meters per minute. how long does it take to walk 2 kilometers?
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What number is 120% of 40?
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3 0
4 years ago
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:

  1. np ≥ 10
  2. 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
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