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3 years ago

HELLLPPPPPP!!! Solve 2x + 2 > 10

Mathematics

1 answer:

olga2289 [7]3 years ago

5 0

(x=5) so 2×5=10+2=12

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Which one is true about the diagram?

In-s [12.5K]

It could be all it could be one but the true answer is I’m not sure which one

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3 years ago

Grace needs $45 to go to Six Flags. She has $13. She earns $8 an hour walking dogs. Which

agasfer [191]

Answer: C. 13 + 8x = 45

Step-by-step explanation:

She already has 13 dollars. This would be the b-value in the equation, thus eliminating options A and B because those two feature 13 as the m value.

She earns $8 per hour. Let x = number of hours walking dogs.

Since 8 is the slope, it would be attached to x, thus either C or D would be the correct response

Since Grace is adding 8 dollars per hour, not subtracting, D is incorrect, leaving C to be the correct answer.

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3 years ago

The American Water Works Association reports that the per capita water use in a single-family home is 63 gallons per day. Legacy

Debora [2.8K]

Answer:

$t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784$

$df=n-1=28-1=27$

$p_v =P(t_{(27)}$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=60$ represent the sample mean

$s=8.9$ represent the sample standard deviation

$n=28$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is less than 63, the system of hypothesis would be:

Null hypothesis: $\mu \geq 63$

Alternative hypothesis: $\mu < 63$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=28-1=27$

Since is a one side lower test the p value would be:

$p_v =P(t_{(27)}$

Conclusion

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4 years ago

.) Find the sum of the first n terms of the geometric sequence for the values of a1and r.

Vinvika [58]

Sn = a(r^n - 1)/(r - 1)
S4 = 228(4.8^4 - 1)/(4.8 - 1) = 228(530.8416 - 1)/3.8 = 228(529.8416)/3.8 = 120,803.8848 / 3.8 = 31,790.496

3 0

4 years ago

A subway has good service 70% of the time and runs less frequently 30% of the time because of signal problems. When there are si

nordsb [41]

Answer:

Step-by-step explanation:

Let GS denote the good service and SP denote the signal problem.

A subway has good service 70% of the time, that is, $P(GS)=0.7$ and a subway runs less frequently 30% of the time because of the signal problems, that is, $P(SP)=0.3$ .

If there are signal problems, the amount of time T in minutes that have to wait at the platform is described by the probability density function given below:

$P_{T|SP}(t)=0.1e^{0.1t}$

If there is good service, the amount of time T in minutes that have to wait at the platform is described the probability density function given below:

$P_{T|GOOD}(t)=0.3e^{0.3t}$

(a)

The probability that you wait at least 1 minute if there is good service P(T ≥ 1| GS) is obtained as follows:

$P(T\geq 1|GS)=\int\limits^{\infty}_1 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_1 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_1\\\\=-(0-e^{-0.3})\\=0.74$

(b)

The probability that you wait at least 1 minute if there is signal problems P(T ≥ 1| SP) is obtained as follows:

$P(T\geq 1|SP)=\int\limits^{\infty}_1 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_1 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.3})]\\\\=-(e^{-0.1t})\limits^{\infty}_1\\\\=-(e^{\infty}-e^{-0.1})\\=-(0-0.904)\\=0.904$

(c)

After 1 minute of waiting on the platform, the train is having signal problems follows an

exponential distribution with parameter $\lambda= 0.1$

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is obtained using the result given below:

$P(SP|T\geq 1)=\frac{P(T\geq 1|SP)P(SP)}{P(T\geq 1)}$

$P(T\geq 1|GS)=0.74, P(T\geq 1|SP)=0.904$

Now calculate the $P(T \geq 1)$ as follows:

$P(T \geq 1)=P(T\geq 1|SP)P(SP)+P(T\geq 1|GS)P(GS)\\=(0.904)(0.3)+(0.74)(0.7)=0.7892$

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is calculated as follows:

$P(SP|T\geq 1)= \frac{0.904 \times 0.3}{0.7892}
= 0.3436
$

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is $0.3436$ .

(d)

After 5 minutes of waiting on the platform, the train is having signal problems follows an exponential distribution with parameter $\lambda= 0.1$ .

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is obtained using the result given below:

$P(SP|T\geq 5)=\frac{P(T\geq 5|SP)P(SP)}{P(T\geq 5)}$

First, calculate the $P(T\geq 5|SP)$ as follows:

$P(T\geq 5|SP)=\int\limits^{\infty}_5 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_5 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.1})]\\\\=-(e^{-0.1t})\limits^{\infty}_5\\\\=-(e^{\infty}-e^{-0.5})\\=-(0-0.6065)\\=0.6065$

Now, calculate the $P (T\geq5|GS )$ as follows:

$P(T\geq 5|GS)=\int\limits^{\infty}_5 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_5 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_5\\\\=-(0-e^{-1.5})\\=0.2231$

Now, calculate the $P (T \geq 5)$ as follows:

$P(T \geq 5)=P(T\geq 5|SP)P(SP)+P(T\geq 5|GS)P(GS)\\=(0.6065)(0.3)+(0.2231)(0.7)=0.3381$

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is calculated as follows:

$P(SP|T\geq 5)= \frac{0.6065 \times 0.3}{0.3381}
= 0.5381
$

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is $0.5381$ .

6 0

3 years ago