In the waiting room of a vet's office, there are 4 cats, 5 dogs, 1 lizard, and 1 hamster.

WINSTONCH [101]

Answer:

$- 5 \frac{1}{3} \div 3 \frac{1}{10} \\ = \frac{ - 5 \times 3 + 1}{3} \div \frac{3 \times 10 + 1}{10} \\ = \frac{ - 16}{3} \times \frac{10}{31} \\ = \frac{ - 160}{93}$

Hope it will help :)

8 0

3 years ago

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Multiply negative 1 over 2 multiplied by negative 1 over 2. Which of the following is correct?

krok68 [10]

Answer:

negative 1 over 4

Step-by-step explanation:

6 0

3 years ago

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Find the solutions to the equation -5/2x+3/4x=-3\4

Usimov [2.4K]

Simplify both sides of the equation

-5/2 x +3/4 x = -3/4

Combine like terms

(-5/2 x + 3/4 x )

-7/4 x = -3/4

Multiply both sides by 4/(-7)

(4/-7)*(-7/4 x ) = (4/-7 ) * ( -3/4 )

x = 3/7

6 0

3 years ago

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<img src="https://tex.z-dn.net/?f=%20%7B8%20%7D%5E%7B2%7D%20%20%2B%20%20%7B10%7D%5E%7B3%7D%20%20%5Cdiv%20%20%7B5%7D%5E%7B2%7D%20

Maslowich

Use PEMDAS

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

$8^2+10^3\div5^2=64+1000\div25=64+40=104$

5 0

3 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 X = number of senior professionals who thought that global warming is having a significant impact on the environment.

The random variable X follows a Binomial distribution with parameters n = 100 and p = 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 p 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 z for P (Z < z) = 0.95 is

z = 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 z for P (Z < z) = 0.975 is

z = 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