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3 years ago

Simplify the expression.

Mathematics

1 answer:

Yuri [45]3 years ago

3 0

Answer:

The answer to your question is letter A) cot x

Step-by-step explanation:

$\frac{1 - cot x}{tan x - 1}$

Remember that cot x = $\frac{1}{tan x}$

Then

$\frac{1 - \frac{1}{tan x} }{tanx -1}$

Simplify

$\frac{\frac{tan x - 1}{tan x} }{tan x - 1}$

$\frac{tan x - 1}{tan x (tan x - 1)}$

$\frac{1}{tan x}$

Result

$\frac{1}{tan x} = cot x$

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Solve each system of equations by substitution <br> X + 3y = 6<br> -x + y = -7

Naily [24]

Answer:

(27/4, -1/4)

Step-by-step explanation:

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3 years ago

In the formula A(t) = A0ekt, A(t) is the amount of radioactive material remaining from an initial amount A0 at a given time t an

Alenkasestr [34]

Answer: 693 years

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

$A(t) =A_0e^{kt}$

k = rate constant

t = time taken for decomposition = 1

$A_0$ = Initial amount of the reactant

$A_t$ = amount of the reactant left = $A_0-\frac{0.1}{100}\times A_0=0.999A_0$

$0.999A_0=A_0e^{k\times 1}$

$0.999=e^k$

$k=-0.001year^{-1}$

for half life : $t=t_\frac{1}{2}$

$A_t=\frac{1}{2}A_o$

Putting in the values , we get

$\frac{1}{2}A_0=A_0e^{-0.001\times t_\frac{1}{2}$

$t_\frac{1}{2}=693 years$

Thus half life of this isotope, to the nearest year is 693.

8 0

3 years ago

The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age gro

ra1l [238]

Answer:

a) Null hypothesis: $p\leq 0.12$

Alternative hypothesis: $p > 0.12$

b) $z_{\alpha}=1.64$

c) $z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362$

d) $p_v =P(z>1.362)=0.0866$

e) For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

Step-by-step explanation:

Information given

n=1000 represent the random sample selected

X=134 represent the number of young drivers ages 18 – 24 that had an accident

$\hat p=\frac{134}{1000}=0.134$ estimated proportion of young drivers ages 18 – 24 that had an accident

$p_o=0.12$ is the value that we want to verify

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v{/tex} represent the p valuePart aWe want to verify if the population proportion of young drivers, ages 18 – 24, having accidents is greater than 12%: Null hypothesis:[tex]p\leq 0.12$

Alternative hypothesis: $p > 0.12$

The statistic would be given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Part b

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the right and we got:

$z_{\alpha}=1.64$

Part c

For this case the statistic would be given by:

$z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362$

Part d

The p value can be calculated with the following probability:

$p_v =P(z>1.362)=0.0866$

Part e

For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

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2 years ago

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tino4ka555 [31]

Answer:

The first choice

Explanation:

Because there is a closed dot, you need to use either less than or equal to or greater than or equal to. And because it’s pointing to left and getting smaller, d is less than or equal to 5.

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