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2 years ago

-3(2x+7)=-51 what does x=

Mathematics

1 answer:

devlian [24]2 years ago

3 0

Answer:

x=5

Step-by-step explanation:

-3(2x+7)=-51

opening bracket in the left hand side

-3.2x+(-3).7=-51

-6x-21=-51

adding 21 on both sides

-6x=-51+21

-6x=-30

dividing both sides by -6

[tex]\frac{-6x}{-6}=\frac{-30}{-6}

x=5

Hence the value of x came out to be 5

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Assume that a sample is used to estimate a population proportion μ . Find the margin of error M.E. that corresponds to a sample

hichkok12 [17]

Answer:

$MoE = 1.645\cdot \frac{13.1}{\sqrt{772} } \\\\MoE = 1.645\cdot 0.47147\\\\MoE = 0.776\\\\$

Step-by-step explanation:

Since the sample size is quite large, we can use the z-distribution.

The margin of error is given by

$MoE = z_{\alpha/2}(\frac{s}{\sqrt{n} } )$

Where n is the sample size, s is the sample standard deviation and $z_{\alpha/2}$ is the z-score corresponding to a 90% confidence level.

The z-score corresponding to a 90% confidence level is

Significance level = α = 1 - 0.90= 0.10/2 = 0.05

From the z-table at α = 0.05

z-score = 1.645

$MoE = 1.645\cdot \frac{13.1}{\sqrt{772} } \\\\MoE = 1.645\cdot 0.47147\\\\MoE = 0.776\\\\$

Therefore, the margin of error is 0.776.

3 0

2 years ago

Radioactive Decay:

Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

Answer: The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

Step-by-step explanation:

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}$

The formula used to calculate the time period for a first order reaction follows:

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}$

where,

k = rate constant = $0.01506days^{-1}$

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

$t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days$

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

7 0

2 years ago