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shepuryov
6 days ago
10

What are the domain range and asymptote of h(x)=(1.4)^x+5

Mathematics
2 answers:
Zina [3.9K]6 days ago
8 0

domain: {x | x is a real number}; range: {y | y > 5}; asymptote: y = 5

Inessa [3.9K]6 days ago
6 0
The range consists of all the valid y values, starting from 5.
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<span>Considering the visitor count is likely rounded to the nearest hundred thousand, the precise figure could range from 350,000 to 449,999. If rounded to the nearest ten thousand, it would be between 395,000 and 404,999.</span>
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5 days ago
Joan draws a bar chart to show the temperatures at midday on March 1st in five cities. Joan has made four mistakes in her diagra
tester [3916]

Answer:

Step-by-step explanation:

Characteristics of a bar graph include:

1). There must be uniform spacing between the bars or columns.

2). Each bar or column should have a consistent width.

3). All bars must share the same baseline.

4). The height of each bar corresponds to the data value.

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- Spacing between London-Paris and Rome-Oslo isn’t uniform.

- Width of the Munich bar differs from the others.

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10 days ago
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What is the factored form of the polynomial x2 - 16x + 48
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(x - 12) (x - 4) is the answer

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6 days ago
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The perimeter of a rectangle is 202. The length is 26 more than 4 times the width. Find the dimensions
Inessa [3907]

Answer:

P = 2*(26+4y) + 4y

Upon solving for y, we find:

202= 52 +8y +4y

202= 52 +12 y

By substitution, we have:

150 = 12y

Thus, we arrive at:

y = \frac{150}{12}= 12.5

And for x, we determine:

x = 26 +4*12.5 = 76

This gives us a length of 76 and a width of 12.5.

Step-by-step explanation:

We are dealing with a rectangle. The formula for the perimeter is:

P= 2x+2y

Here, x denotes the length and y the width. The following conditions can be established:

x = 26 +4y

Substituting these values yields:

P = 2*(26+4y) + 4y

When we solve for y, we find:

202= 52 +8y +4y

202= 52 +12 y

By substitution, we obtain:

150 = 12y

Thus, we conclude:

y = \frac{150}{12}= 12.5

The length is found to be 76, while the width measures 12.5.

6 0
4 days ago
Two random samples are taken from private and public universities
babunello [3635]

Response:

Detailed explanation:

For private institutions,

n = 20

Average, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05

Standard deviation = √(sum of (x - mean)²/n

Sum of (x - mean)² = (43120 - 34623.05)^2 + (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95

Standard deviation = √(1527829234.95/20

s1 = 8740.22

For public institutions,

n = 20

Average, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15

Sum of (x - mean)² = (25469 - 25063.15)^2 + (19450 - 25063.15)^2 + (18347 - 25063.15)^2 + (28560 - 25063.15)^2 + (32592 - 25063.15)^2 + (21871 - 25063.15)^2 + (24120 - 25063.15)^2 + (27450 - 25063.15)^2 + (29100 - 25063.15)^2 + (21870 - 25063.15)^2 + (22650 - 25063.15)^2 + (29143 - 25063.15)^2 + (25379 - 25063.15)^2 + (23450 - 25063.15)^2 + (23871 - 25063.15)^2 + (28745 - 25063.15)^2 + (30120 - 25063.15)^2 + (21190 - 25063.15)^2 + (21540 - 25063.15)^2 + (26346 - 25063.15)^2 = 1527829234.95

Standard deviation = √(283738188.55/20

s2 = 3766.55

This involves two independent samples. Define μ1 as the mean out-of-state tuition for private institutions and μ2 as the mean out-of-state tuition for public institutions.

The random variable represents μ1 - μ2 = the difference between the mean out-of-state tuition for private vs. public institutions.

The hypothesis is established as follows. The correct choice is

-B. H0: μ1 = μ2; H1: μ1 > μ2

As the sample standard deviation is known, the test statistic is calculated using the t test formula:

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)

t = 9559.9/2128.12528473889

t = 4.49

The method for finding degrees of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114

df = 26

The probability value is obtained from the t test calculator. It is

p value = 0.000065

Given that alpha, 0.01 > the p value, 0.000065, we will reject the null hypothesis. Hence, at a significance level of 1%, the mean out-of-state tuition for private institutions is statistically significantly greater than that of public institutions.

4 0
7 days ago
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