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3 years ago

Identify Slope and Y intercept for the equation X=4 M=B=

1 answer:

6 0

Undefined and Undefined

Since the equation is x = 4, there is no y-intercept as x will never equal 0.

Additionally, since the equation is x = 4, there is no change in x value and as such the slope can not be determined.

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Calculate the income needed based on the following budgeted expenses.

Lynna [10]

Hope this helped you :)

5 0

3 years ago

Simplify x^3y^4/3y^4

yan [13]

The given expression is:
(x^3 * y^4) / (3y^4)

We can notice that the term y^4 is a common term found in both the numerator and the denominator of the given expression, therefore, we can cancel this term from the both numerator and denominator (as if you divided both numerator and denominator by y^4).

Doing this, we will have the simplified form of the expression as follows:
(x^3) / (3)

4 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Anettt [7]

Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be $X$ centimeters.

$X$ follows a normal distribution with

mean $\mu = 11.2$ and
standard deviation $\sigma = 2.1$ .

In other words,

$X\sim \text{N}(11.2, 2.1^{2})$ ,

and the probability in question is $X \le 13$ .

Z-score table approach:

Find the z-score of this measurement:

$\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143$ . Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that $Z$ will be less than or equal to the z-score in question. (In case the question is asking for the probability that $Z$ is greater than the z-score, subtract the value from table from 1.)

$P(X\le 13) = P(Z \le 0.857143) \approx 0.8051$ .

"Technology" Approach

Depending on the manufacturer, the steps generally include:

Locate the cumulative probability function (cdf) for normal distributions.
Enter the lower and upper bound. The lower bound shall be a very negative number such as $-10^{9}$ . For the upper bound, enter $13$
Enter the mean and standard deviation (or variance if required).
Evaluate.

For example, on a Texas Instruments TI-84, evaluating $\text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 )$ gives $0.804317$ .

As a result,

$P(X\le 13) = 0.804317$ .

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

Number of trials $n$ : 150.
Probability of success $p$ : 0.804317.

Let $Y$ be the number of green beans out of this 150 that are shorter than 13 centimeters. $Y\sim\text{B}(150,0.804317)$ .

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

$E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121$

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

7 0

3 years ago

A survey of 2,254 American adults indicates that 17% of cell phone owners browse the internet exclusively on their phone rather

GrogVix [38]

Answer:

We conclude that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%.

Step-by-step explanation:

We are given that a survey of 2,254 American adults indicates that 17% of cell phone owners browse the internet exclusively on their phone rather than a computer or other device. According to an online article, a report from a mobile research company indicates that 38 percent of Chinese mobile web users only access the internet through their cell phones.

We have to conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%.

Let p = proportion of Americans who only use their cell phones to access the internet

SO, Null Hypothesis, $H_0$ : p = 38% {means that the proportion of Americans who only use their cell phones to access the internet is same as that of Chinese proportion of 38%}

Alternate Hypothesis, $H_a$ : p $\neq$ 38% {means that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of cell phone owners who browse the internet

exclusively on their phone in a survey of 2,254 adults = 17%

n = sample of adults = 2,254

So, test statistics = $\frac{0.17-0.38}{\sqrt{\frac{0.17(1- 0.17)}{2,254} } }$

= -26.542

Since in the question we are not given with the significance level so we assume it to be 5%. So, at 0.05 level of significance, the z table gives critical values between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie in between the critical values of z so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%.

3 0

3 years ago

Log 3=.4771 log 5=.6990 find the value of log 150

raketka [301]

Answer:

2.17609

Step-by-step explanation:

Easiest and fastest way is to just directly plug log base 10 of 150 into the calc, as it is a nasty decimal.

4 0

3 years ago