All categories

3 years ago

PLEASE HELP!! WILL MARK BRAINLIEST :))

Mathematics

1 answer:

igor_vitrenko [27]3 years ago

5 0

Answer:

It would be the third one (x, y) for all real numbers

You might be interested in

E right 75 80 85 90 95 100 105 110 115 120 125 130 135 What is the median

amid [387]

105 is ur answer for this question

3 0

4 years ago

Read 2 more answers

PLEASE HELP ALGEBRA 2!!

julsineya [31]

Answer:

11) 2nd option

12)2nd option

13)3rd option

8 0

4 years ago

Helpppppppppppppppppppp mathhhhhhhhh

sp2606 [1]

2) angle T Congruent to angle D; 3)angle s and angle u

7 0

3 years ago

If -7.6(x - 1) = 7.3 - 7.4x, find the value of -2x.

Marat540 [252]

Answer:

-3

Step-by-step explanation:

Step 1: Write equation

-7.6(x - 1) = 7.3 - 7.4x

Step 2: Solve for x

Distribute -7.6: -7.6x + 7.6 = 7.3 - 7.4x

Add 7.6x on both sides: 7.6 = 7.3 + 0.2x

Subtract 7.3 on both sides: 0.3 = 0.2x

Divide both sides by 0.2: x = 1.5

Step 3: Find -2x

-2(1.5) = -3

5 0

4 years ago

Read 2 more answers

Suppose the FAA weighed a random sample of 20 airline passengers during the summer and found their weights to have a sample mean

Orlov [11]

Answer:

Step-by-step explanation:

Given Parameters

Mean, $x$ = 180

total samples, n = 20

Standard dev, $\sigma$ = 30

$\alpha$ = 1 - 0.95 = 0.05 at 95% confidence level

Df = n - 1 = 20 - 1 = 19

Critical Value, $t_\alpha$ , is given by

$t_{c}=t_{\alpha, df} = t_{0.05,19} = 1.729$

a).

Confidence Interval, $\mu$ , is given by the formula

$\mu = x +/- t_c \times \frac{s}{\sqrt{n} }$

$\mu = 180 +/- 1.729 \times \frac{30}{\sqrt{20} }$

$\mu = 180 +/-11.5985$

$191.5985 > \mu > 168.4015$

b).

Critical Value, $t_{\alpha/2}$ , is given by

$t_{c}=t_{\alpha/2, df} = t_{0.05/2,19} = 2.093$

Confidence Interval, $\mu$ , is given by

$\mu = x +/- t_c \times \frac{s}{\sqrt{n} }$

$\mu = 180 +/- 2.093 \times \frac{30}{\sqrt{20} }$

= 180 +/- 14.0403

= 165.9597 < $\mu$ < 194.0403

8 0

3 years ago