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3 years ago

Kayla walks 4 2/5 miles each day.

Mathematics

2 answers:

kow [346]3 years ago

8 0

I think the answer is b

Zarrin [17]3 years ago

4 0

Yeah the answer is B

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Write an equation with two variables that represents a real word situation

KATRIN_1 [288]

Solve for t. d = rt heres a real word question. but im not sure if you can use this exact one :)

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4 years ago

The sides of a square are three to the power of two sevenths inches long. What is the area of the square? (5 points) nine to the

kakasveta [241]

It's three to the power of four sevenths square inches.
Hope this will help you!

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3 years ago

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Determine whether the graph of each pair of equations are parrallel, perpendicular, or neither

Luba_88 [7]

Answer: 1: parrallel

2:perpendicular

3; neither

4; perpendicular. 5: parrallel

Step-by-step explanation:

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3 years ago

Find the dimensions of a rectangle whose perimeter is 16 meters and whose area is 15 square meters.

Veseljchak [2.6K]

2(x+y)=16,
x+y=16/2=8
y=8-x
x*y=15
x(8-x)=15
8x-x^2=15
x^2-8x+15=0
x^2-3x-5x+15=0
x(x-3)-5(x-3)=0
(x-3)(x-5)=0
x=3,5
when x=3
y=8-3=5
when x=5
y=8-5=3
length=5 meters
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3 0

3 years ago

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A researcher has developed a new drug designed to reduce blood pressure. In an experiment, 21 subjects were assigned randomly to

Paladinen [302]

Answer:

Step-by-step explanation:

Hello!

The objective of the research is to compare the newly designed drug to reduce blood pressure with the standard drug to test if the new one is more effective.

Two randomly selected groups of subjects where determined, one took the standard drug (1- Control) and the second one took the new drug (2-New)

1. Control

X₁: Reduction of the blood pressure of a subject that took the standard drug.

n₁= 23

X[bar]= 18.52

S= 7.15

2. New

X₂: Reduction of the blood pressure of a subject that took the newly designed drug.

n₂= 21

X[bar]₂= 23.48

S₂= 8.01

The parameter of study is the difference between the two population means (no order is specified, I'll use New-Standard) μ₂ - μ₁

Assuming both variables have a normal distribution, there are two options to estimate the difference between the two means using a 95% CI.

1) The population variances are unknown and equal:

[(X[bar]₂-X[bar]₁)± $t_{n_1+n_2-2;1-\alpha /2}$ *( $Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ )]

$t_{n_1+n_2-2;1-\alpha /2}= t_{23+21-2;1-0.025}= t_{42;0.975}= 2.018$

$Sa=\sqrt{\frac{(n_1-1)*S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{22*7.15^2+20*8.01^2}{42} }= 7.57$

[23.48-18.52]±2.018*( $7.57*\sqrt{\frac{1}{21} +\frac{1}{23} }$ )]

[0.349; 9.571]

2) The population variables are unknown and different:

Welche's approximation:

[(X[bar]₂-X[bar]₁)± $t_{Dfw;1-\alpha /2}$ *( $\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} }$ )]

$Df_{w}= \frac{(\frac{S_1^2}{n_1} +\frac{S^2_2}{n_2} )^2}{\frac{(\frac{S_1^2}{n_1} )^2}{n_1-1}+ \frac{(\frac{S_2^2}{n_2} )^2}{n_2-1} } = \frac{(\frac{7.15^2}{23} +\frac{8.01^2_2}{21} )^2}{\frac{(\frac{7.15^2}{21} )^2}{20}+ \frac{(\frac{8.01^2}{23} )^2}{22} } = 42.85= 42$

$t_{Df_w;1-\alpha /2}= t_{42; 0.975}= 2.018$

[(23.48-18.52)±2.018 $\sqrt{\frac{7.15^2}{23} +\frac{8.01^2}{21} }$ ]

[0.324; 9.596]

I hope this helps!

4 0

3 years ago