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3 years ago

the sum of a negative number, ¼ of the negative number, and 7/16 off the negative number = -13½. what is the negative number?

1 answer:

6 0

X - the number

$x+\frac{1}{4}x+\frac{7}{16}x=-13 \frac{1}{2} \\
\frac{16}{16}x+\frac{4}{16}x+\frac{7}{16}x=-\frac{27}{2} \\
\frac{27}{16}x=-\frac{27}{2} \\
x=-\frac{27}{2} \times \frac{16}{27} \\
x=-\frac{1}{2} \times \frac{16}{1} \\
x=-8$

The number is -8.

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Find the indicated probability or percentage for the sampling error. The distribution of weekly salaries at a large company is r

Flauer [41]

Answer:

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 80 employees will be at most $75 is 0.9297.

Step-by-step explanation:

According to the Central Limit Theorem if we have a non-normal population with mean μ and standard deviation σ and appropriately huge random samples (n > 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 means is given by,

$\mu_{\bar x}=\mu$

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

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

The information provided is:

$\mu=\$1000\\\sigma=\$370\\n=80$

As n = 80 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean weekly salaries.

Let $\bar X$ represent the sample mean weekly salaries.

The distribution of $\bar X$ is: $\bar X\sim N(\$1000,\ \$41.37)$

Now we need to compute the probability of the sampling error made in estimating the mean weekly salary to be at most $75.

The sampling error is the the difference between the estimated value of the parameter and the actual value of the parameter, i.e. in this case the sampling error is, $|\bar X-\mu|= 75$ .

Compute the probability as follows:

$P(-75$

$=P(-1.81$

Thus, 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 80 employees will be at most $75 is 0.9297.

3 0

3 years ago

What is the probability of getting at least three heads when flipping four coins?

umka21 [38]

0.25% chance of getting at least three heads out of four coin flips.

8 0

3 years ago

Combine like terms to simplify the equation below.

Bas_tet [7]

Answer:

4a+6b

Step-by-step explanation:

4a -2a +6b +2a

Combine like terms

4a-2a+2a+6b

4a+6b

6 0

3 years ago

Read 2 more answers

Maple Grove wants to include two typesof maple trees, the namesake of thecity, for their parks. Two varieties ofmaple trees have

blsea [12.9K]

X = number of $85 trees, y = number of $95 trees

Logically, we can construct the expression 85x + 95y to represent how much Maple Grove spends on their trees in total. They want this cost to be less than or equal to $95,000. This means that the inequality has to be 85x + 95y ≤ 95000, either choice a or b. Since the community wants more $95 trees, they want y to be greater than or equal to x.

This means choice b: 85x + 95y ≤ 95000, x≤y is your answer.

Hope this helps!

8 0

3 years ago

In a random sample of 10 residents of the state of Washington, the mean waste recycled per person per day was 2.3 pounds with a

djverab [1.8K]

Answer:

90% confidence interval for the mean waste recycled per person per day for the population of Washington is [2.074 , 2.526].

Step-by-step explanation:

We are given that a a random sample of 10 residents of the state of Washington, the mean waste recycled per person per day was 2.3 pounds with a standard deviation of 0.39 pounds.

Firstly, the pivotal quantity for 90% confidence interval for the mean waste recycled per person per day for the population of Washington is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean waste recycled per person per day = 2.3 pounds

s = sample standard deviation = 0.39 pounds

n = sample of residents = 10

$\mu$ = population mean waste recycled per person per day

Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.833 < $t_9$ < 1.833) = 0.90 {As the critical value of t at 9 degree of

freedom are -1.833 & 1.833 with P = 5%}

P(-1.833 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.833) = 0.90

P( $-1.833 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.833 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.833 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.833 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $2.3-1.833 \times {\frac{0.39}{\sqrt{10} } }$ , $2.3-1.833 \times {\frac{0.39}{\sqrt{10} } }$ ]

= [2.074 , 2.526]

Therefore, 90% confidence interval for the mean waste recycled per person per day for the population of Washington is [2.074 , 2.526].

8 0

3 years ago