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iVinArrow [24]
3 years ago
5

Quincy has $37 in a savings account​

Mathematics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

The 2nd answer would be the correct option.

Step-by-step explanation:

Hope this helps:) Also can I have brainliest if possible:) It is ok if you can not.

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Two rigid transformations are used to map δhjk to δlmn. The first is a translation of vertex h to vertex l. What is the second t
Natalka [10]

Answer:

B a rotation about point H

Step-by-step explanation:

Two rigid transformations are used to map ΔHJK to ΔLMN. The first is a translation of vertex H to vertex L. What is the second transformation?

a reflection across the line containing HK

a rotation about point H

a reflection across the line containing HJ

a rotation about point K

8 0
3 years ago
Read 2 more answers
An economist uses the price of a gallon of milk as a measure of inflation. She finds that the average price is $3.82 per gallon
salantis [7]

Answer:

(a) The standard error of the mean in this experiment is $0.052.

(b) The probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c) The probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d) The likelihood that the sample mean is greater than $3.92 is 0.9726.

Step-by-step explanation:

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

n=40\\\mu=\$3.82\\\sigma=\$0.33

As <em>n</em> = 40 > 30, the distribution of sample mean is \bar X\sim N(3.82,\ 0.052^{2}).

(a)

The standard error is the standard deviation of the sampling distribution of sample mean.

Compute the standard deviation of the sampling distribution of sample mean as follows:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

    =\frac{0.33}{\sqrt{40}}\\\\=0.052178\\\\\approx 0.052

Thus, the standard error of the mean in this experiment is $0.052.

(b)

Compute the probability that the sample mean is between $3.78 and $3.86 as follows:

P(3.78

                               =P(-0.77

Thus, the probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c)

If the difference between the sample mean and the population mean is less than $0.01 then:

\bar X-\mu_{\bar x}

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

Thus, the probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d)

Compute the probability that the sample mean is greater than $3.92 as follows:

P(\bar X>3.92)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{3.92-3.82}{0.052})

                    =P(Z

Thus, the likelihood that the sample mean is greater than $3.92 is 0.9726.

3 0
3 years ago
Solve 10e^2x - 5 =23^x for x.
yawa3891 [41]

10e^{2x}-5=23e^x\qquad\text{subtract}\ 23e^x\ \text{from both sides}\\\\10e^{2x}-23e^x-5=0\\\\10(e^x)^2-23(e^x)-5=0\\\\\text{substitution:}\ e^x=t > 0\\\\10t^2-23t-5=0\\\\10t^2-25t+2t-5=0\\\\5t(2t-5)+1(2t-5)=0\\\\(2t-5)(5t+1)=0\iff2t-5=0\ \vee\ 5t+1=0\\\\2t=5\ \vee\ 5t=-1\\\\t=\dfrac{5}{2} > 0\ \vee\ t=-\dfrac{1}{5} < 0\\\\\text{therefore}\ e^x=\dfrac{5}{2}\to\ln e^x=\ln\left(\dfrac{5}{2}\right)\\\\\boxed{x=\ln\left(\dfrac{5}{2}\right)}

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3 years ago
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2(g+5)=-7-4g+g include instructions plz!
Elza [17]
First distribute the 2, then add 7 to that side, so you have 2g+17=-4g+g , subtract 2g and it should be easy to find from there lol
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I have a water tank that holds 5 gallons of water I drink 4 pints of water a day how many full tanks of water I will be drinking
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