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3 years ago

If a country’s dept to gdp ratio is 142% the country is borrowing more than it is producing.

Mathematics

2 answers:

CaHeK987 [17]3 years ago

7 0

True because if the country's GDP was less than 100%, it would be losing money instead of making money.

JulsSmile [24]3 years ago

5 0

Answer:

A. True

Step-by-step explanation:

In the question it is given that the dept to GDP ratio is 142%. This mean that

$\frac{Dept}{GDP} = \frac{142}{100}$

this can be interpreted as when GDP is 100, the dept on the nation is 142. Clearly, dept is more than the GDP of the country. This means that the country is borrowing more than it is producing. Hence, the given statement is true

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In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

Write the equation of the line, given the y- and x-intercepts x-intercept (–6, 0), y-intercept (0, –8) PLLEEEEEZE HEEELP

ad-work [718]

Answer is
Y= -4/3x-8

7 0

3 years ago

A micro-photograph of a human hair shows the hair magnified 150 times. If the hair is

stepan [7]

Answer:

The actual width of the hair is 20 / 150 = 0.13 mm

Step-by-step explanation:

Scaling

Objects can be represented in a reduced or augmented size by using scaling which is basically multiplying or dividing by a constant factor. We use scaling in several applications of real-life.

The human hair is magnified 150 times by a micro-phtograph. This is the scale factor.

The hair can be seen 2 cm wide in the photograph.

Converting to millimeters, 2 cm = 2*10 = 20 millimeters.

The actual width of the hair is 20 / 150 = 0.13 mm

6 0

2 years ago

What is the solution to the following equation? 3(X - 4)- 5 = X - 3 O A. X = 12 B. X = 8 O C. x=7 O D. *= 3

Andrews [41]

Answer: c.) x = 7

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Find the average rate of change of k(x)=5x–19 over the interval [ - 14, - 4].

rusak2 [61]

The rate of change of a function can be modeled with the following expression:

$\frac{\Delta{k{x)}}{\Delta{x}}$

Where Δx is the change in x value, and Δk(x) is the corresponding change in k(x). We're given the two extremes of x, so we can calculate the change in x to be

$\Delta{x}=-4-(-14)=10$

To find the change in k(x), we can calculate the values of k(x) at x = -14 and x = -4 and find the difference between them:

$k(-14)=5(-14)-19=-70-19=-89\\ k(-4)=5(-4)-19=-20-19=-39\\ \Delta{y}=k(-14)-k(-4)=-89-(-39)=-50$

So, the rate of change for the function from x = -14 to x = -4 is

$\frac{\Delta{y}}{\Delta{x}}= \frac{-50}{10}=-5$

7 0

3 years ago