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3 years ago

1×+3=4×-6need help finding this out

Mathematics

2 answers:

stellarik [79]3 years ago

7 0

1x+3 =4x-6

subtact 1x from each side

3 = 3x -6

add 6 to each side

9 = 3x

divide each side by 3

3=x

sladkih [1.3K]3 years ago

3 0

Answer:

x = 3

Step-by-step explanation:

x + 3 = 4x - 6

add 6 to both sides

x + 9 = 4x

subtract x from both sides

9 = 3x

divide 3 from both sides

x = 3

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Before the distribution of certain statistical software, every fourth compact disk (CD) is tested for accuracy. The testing proc

lilavasa [31]

Answer:

a) For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

$P(R) =\frac{1}{4} =0.25$

The failure probability for each program are given by:

$P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01$

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

$P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932$

So then the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

And if we use the fact that we have 4 possible test the true probability of interest would be:

$P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017$

b) $p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))$

And replacing we got:

$p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484$

c) From part a we now that the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

So then if we have 100 CDs the expected number of rejected Cd's are:

$100*0.068= 6.8 \approx 7$

Step-by-step explanation:

Part a

For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

$P(R) =\frac{1}{4} =0.25$

The failure probability for each program are given by:

$P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01$

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

$P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932$

So then the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

And if we use the fact that we have 4 possible test the true probability of interest would be:

$P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017$

Part b

For this case we want the probability that it failed program 2 or 3

So then we can find this probability like this:

$p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))$

And replacing we got:

$p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484$

Part c

From part a we now that the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

So then if we have 100 CDs the expected number of rejected Cd's are:

$100*0.068= 6.8 \approx 7$

3 0

3 years ago

What is m∠2? ° What is m∠1? °

stepan [7]

What is m∠2 = m∠6 = 75

What is m∠1 = 180 - 75 = 105

7 0

3 years ago

Read 2 more answers

A manufacturer of metal fasteners expects to ship an average of 1197.00 boxes of fasteners per day. A random sample of 24 days p

lana66690 [7]

Answer:

The p-value obtained is less than the significance level at which the test was performed, hence, we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Step-by-step explanation:

For hypothesis testing, we first clearly state our null and alternative hypothesis.

The null hypothesis is that there is no evidence showing that the average number of boxes shipped per day is different from 1197.00. That is, there is no significant difference between the number of boxes shipped per day and 1197.00

And the alternative hypothesis is that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Mathematically, the null hypothesis is

H₀: μ₀ = 1197.00

The alternative hypothesis is

Hₐ: μ₀ ≠ 1197.00

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 1193.69 boxes per day

μ₀ = standard that we're comparing the sample mean with = 1197.00

σₓ = standard error of the sample mean

= (σ/√n)

where n = Sample size = 24 days

σ = standard deviation of the sample = 3.3166 boxes per day

σₓ = (3.3166/√24) = 0.677

t = (1193.69 - 1197) ÷ 0.677

t = -4.89

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 24 - 1 = 23

Significance level = 0.05

p-value (for t = -4.89, at 0.05 significance level, with a two tailed condition) = 0.000061

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 5% = 0.05

p-value = 0.000061

0.000061 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Hope this Helps!!!

6 0

3 years ago

How to calculate Relative frequency

Andreyy89

To find the relative frequency, divide the frequency by the total number of data values. To find the cumulative relative frequency, add all of the previous relative frequencies to the relative frequency for the current row.

6 0

3 years ago

Read 2 more answers

NEED HELP FAST

Natasha_Volkova [10]

Your answer is C :) blessings

5 0

3 years ago