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2 years ago

my number has a tens digit that is 8 more than the ones digit zero is not one of my digits my number is

Mathematics

1 answer:

liq [111]2 years ago

8 0

It should be 91 but it varies. but 91 is the least number you can make

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Mrs. Junkin wrote the function f(x) 2/3x -5 on the chalkboard. What is the value of this function for f(6)?

Rama09 [41]

Answer:

-1

Step-by-step explanation:

f(x) =2/3x -5

Let x = 6

f(6) = 2/3 (6) - 5

Multiply first

f(6) = 12/3 -5

= 4-5

= -1

3 0

3 years ago

Find the slope between the two points: (9,9) and (12,15)

grandymaker [24]

The slope formula is $\frac{Y2-Y1}{X2-X1}$ . For the two points, $(x,y)$ . You can pick any of the Y and X from these two points, but only once. $\frac{9-15}{9-12}$ = $\frac{-6}{-3}$ = 2 or $\frac{15-9}{12-9}$ = $\frac{6}{3}$ = 2 and both will equal to. The answer for the slope is 2.

3 0

2 years ago

Read 2 more answers

Pls helpppppppppppppo

vovikov84 [41]

Answer:

b I do believe

Step-by-step explanation:

can you help me

8 0

3 years ago

The formula a = 118e0.024t models the population of a particular city, in thousands, t years after 1998. when will the populatio

andriy [413]

Given that the population has been modeled by the formula:
a=118e^(0.024t), the time taken for the population to hit 140k will be given by:
140000=118e^(0.024t)
solving for t we shall have:
140000/118=e^(0.024t)
thus;
0.024t=ln(140000/118)
t=1/0.024*ln(140000/118)
t=295
thus the time the population will be 140000 will be:
1998+295
=2293

6 0

2 years ago

You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the

il63 [147K]

Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that $n = 2809, \pi = \frac{1634}{2809} = 0.5817$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034$

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

6 0

2 years ago