Need help!!<br> Need to solve this sheet that is attached<br> Please help! 100 points

<img src="https://tex.z-dn.net/?f=2x-y%3D4%5C%5Cy-x%3D4" id="TexFormula1" title="2x-y=4\\y-x=4" alt="2x-y=4\\y-x=4" align="absmi

Flura [38]

Here is the answer. Hope this helps!

5 0

1 year ago

Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

nordsb [41]

$\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\
y = t + 6\ln|t| + C. \text{ But $t\ \textgreater \ 0$ so }y = t + 6\ln t + C. \\ \\ 
y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\
y(t) = t + 6\ln t + 7$

5 0

3 years ago

Can somebody help me simplify these problems please?

Mazyrski [523]

1: answer: 4^6 ( reduce the fraction with 4^3)

2: answer: 1 ( any nonzero expression raised to the power of 0 equals 1 )

3: answer 3^8
steps: (3^2)^4 = 3^2×4 ( to raise a power to another Power multiply the exponents)
3^2×4 = 3^8 ( multiply the 2 and the 4 exponent to get a 8 as a exponent )

7 0

2 years ago

Can anyone help me with this one please:)

mina [271]

Answer:

Look below

Step-by-step explanation:

So you know a point and the slope, now you can substitute the values of the point for the equation.

5 0

2 years ago

Read 2 more answers

The distribution of weekly salaries at a large company is right skewed with a mean of $1000 and a standard deviation of $350. Wh

Shtirlitz [24]

Answer:

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 1000, \sigma = 350, n = 50, s = \frac{350}{\sqrt{50}} = 49.5$

What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

This is the pvalue of Z when X = 1000+50 = 1050 subtracted by the pvalue of Z when X = 1000-50 = 950. So

X = 1050

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1050-1000}{49.5}$

$Z = 1.01$

$Z = 1.01$ has a pvalue of 0.8438

X = 950

$Z = \frac{X - \mu}{s}$

$Z = \frac{950-1000}{49.5}$

$Z = -1.01$

$Z = -1.01$ has a pvalue of 0.1562

0.8438 - 0.1562 = 0.6876

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

4 0

2 years ago

Need help!! Need to solve this sheet that is attached Please help! 100 points