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3 years ago

Sole and check 5y + 7 = 14y + 7

2 answers:

8 0

$5y + 7 = 14y + 7 \\ \\ 5y = 14y \ / \ cancel \ 7\\ \\ 5y - 14y = 0 \ / \ terms \ to \ one \ side\\ \\ -9y = 0 \ / \ simplify \\ \\ y = 0 \ / \ divide \ by \ -9 \\ \\ CHECK \\ \\ 5y + 7 = 14y + 7 \ / \ let \ y = 0 \\ \\ 5 \times 0 + 7 = 14 \times 0 + 7 \\ \\ Answer: \fbox {y = 0}$

MA_775_DIABLO [31]3 years ago

7 0

5y + 7=14y+7
-5 -5
7=9y+7
-7 -7
0 9y
0/9=0
and i dont know if its right and i dont know how to check it

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Arada [10]

Answer: 0.14

Step-by-step explanation:

fourteen nundredths = 0.14

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4 years ago

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I always get lost what is 1-1=. Is it 0 or 1.??????

koban [17]

Answer:

1-1=0 haha

Step-by-step explanation:

if you have one ball and then take away one ball, you will have zero balls

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3 years ago

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

Cube root of - 1000p^12q^3

aleksley [76]

$\bf \sqrt[3]{-1000p^{12}q^3}~~
\begin{cases}
1000=10^3\\
(-1)^3=-1\cdot -1\cdot -1\\
\qquad -1\\
p^{12}=p^{4\cdot 3}\\
\qquad (p^4)^3
\end{cases}\implies \sqrt[3]{(-1)^3(10)^3(p^4)^3q^3}
\\\\\\
(-1)(10)p^4q\implies -10p^4q$

7 0

3 years ago

How many pairs of whole numbers have a sum of 40

Ahat [919]

It is 20
39+1
38+2
37+3
36+4
35+5
<span>Until you get to 20+20</span>

3 0

3 years ago