Find the 9th term of the geometric sequence 10, 40, 160, ...

Pentagon jklmn is translated five units up. what is the name of the image after translation?

mrs_skeptik [129]

There will not be any specific name given to that image after translation

Step-by-step explanation:

Whatever may be the transformation, either it may be rotation, reflection or translation, the size or shape of the image will not change no matter what the condition is and hence the name of that image will also not change and remains constant.
Thus we can conclude that the name of the image will always tend to remain constant and not changes.

4 0

3 years ago

3.2 feet = ____ meters. (3.3 ft = 1 meters)

vodka [1.7K]

Answer:

0.97536

Step-by-step explanation:

Simple, really. Feet to meter conversion apps are on the web.

7 0

3 years ago

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

(c) 90.53%

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

4 years ago

Suppose it is known that 60% of radio listeners at a particular college are smokers. A sample of 500 students from the college i

vladimir1956 [14]

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable X be defined as the number of students at a particular college who are smokers

The random variable X follows a Binomial distribution with parameters n = 500 and p = 0.60.

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 X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

$np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10$

Thus, a Normal approximation to binomial can be applied.

So,

$X\sim N(\mu=600, \sigma=\sqrt{120})$

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

= P (X > 280.50)

$=P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

8 0

3 years ago

Number ten please thanks

BlackZzzverrR [31]

-3^2= -(3)^2

The exponent of 2 only applies to the number 3. -(3)^2 should equal -9. This is true because according to the order of operations, exponents should be evaluated before multiplication. The negative sign here represents -1* 3^2.
If you want to find -3 to the power of 2 it must be written (-3)^2.

5 0

3 years ago

Find the 9th term of the geometric sequence 10, 40, 160, ... ​

Find the 9th term of the geometric sequence 10, 40, 160, ...