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Dovator [93]
3 years ago
13

Line segment AB has endpoints A(1, 4) and B(2, 8) . A dilation, centered at the origin, is applied to AB¯¯¯¯¯ . The image has en

dpoints A′(18, 12) and B′(14, 1) . What is the scale factor of the dilation? 1/8 1/2 2 8
Mathematics
2 answers:
Kruka [31]3 years ago
8 0

Answer:

The answer is 1/8!

Step-by-step explanation:

I took the test and got this question correct! Good luck everyone, you'll do great! Especially with this correct answer =)

kumpel [21]3 years ago
7 0
To determine the scale factor of the dilation, we determine the distances between the endpoints of the two lines through the equation,
                            d = √(x₂ - x₁)² + (y₂ - y₁)²
 Substituting the known values.

Line Segment 1:     d = √(1 - 2)² + (4 - 8)²  = 4.123

Line Segment 2:     d = √(18 - 14)² + (12 - 1)² = 11.70

Dividing the answers will give us 0.352
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One of your friends is testing the effect of drinking coffee on the duration of cold symptoms. The common cold lasts, on average
Nonamiya [84]

Answer:

0.0930

Step-by-step explanation:

Given that one of your friends is testing the effect of drinking coffee on the duration of cold symptoms. The common cold lasts, on average, 6 days.

she tests a one-sided alternative about the population mean cold duration when drinking coffee, H0: μcoffee = 6 Ha: μcoffee < 6 She finds z = −1.68 with one-sided P-value P = 0.0465.

If hypotheses changed to two tailed as:

H_0: \mu_{coffee}  = \mu_{tea}\\H_a: \mu_{coffee}\neq \mu_{tea}

(Two tailed test)

Since we have mean difference and std error the same, test statistic would remain the same.

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6 0
3 years ago
Assume that f(x)=ln(1+x) is the given function and that Pn represents the nth Taylor Polynomial centered at x=0. Find the least
WINSTONCH [101]

Answer:

the least integer for n is 2

Step-by-step explanation:

We are given;

f(x) = ln(1+x)

centered at x=0

Pn(0.2)

Error < 0.01

We will use the format;

[[Max(f^(n+1) (c))]/(n + 1)!] × 0.2^(n+1) < 0.01

So;

f(x) = ln(1+x)

First derivative: f'(x) = 1/(x + 1) < 0! = 1

2nd derivative: f"(x) = -1/(x + 1)² < 1! = 1

3rd derivative: f"'(x) = 2/(x + 1)³ < 2! = 2

4th derivative: f""(x) = -6/(x + 1)⁴ < 3! = 6

This follows that;

Max|f^(n+1) (c)| < n!

Thus, error is;

(n!/(n + 1)!) × 0.2^(n + 1) < 0.01

This gives;

(1/(n + 1)) × 0.2^(n + 1) < 0.01

Let's try n = 1

(1/(1 + 1)) × 0.2^(1 + 1) = 0.02

This is greater than 0.01 and so it will not work.

Let's try n = 2

(1/(2 + 1)) × 0.2^(2 + 1) = 0.00267

This is less than 0.01.

So,the least integer for n is 2

7 0
4 years ago
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