Ask question

Ask question

All categories

3 years ago

13

An employee worked 160.5 in may, 165.75 hours in June and 152.25 hours in July. What was the average number of hours he worked e

ach month?

2 answers:

topjm [15]3 years ago

7 0

To find it use 160.5+165.75+152.25=x then do x/3. Find x yourself and you can solve.

zepelin [54]3 years ago

4 0

239.25 that is your answer

You might be interested in

25(y+1)-x^2(y+1) i don't know the answer

natita [175]

It seems like you've begun to group and factored out the GCF.

Because the inner binomials (the ones in parentheses) are the same, the next step is to rewrite the equation like so:

(25-x^2)(y+1)

From here, we can further factor (25-x^2) since they are both perfect squares.

(5+x)(5-x)(y+1)

No further terms can be factored so that is the final answer. Hope this helps!

7 0

3 years ago

For a particular event, 771 tickets were sold for a total of $3037. If students paid $3 per ticket and non-students paid $5

BlackZzzverrR [31]

Answer:

Number of student tickets were sold = 409

Step-by-step explanation:

Total number of tickets sold = 771

I.e Students ticket (S) + Non students ticket(N S) = 771

Total amount to be paid for tickets = $ 3037

Amount paid by students per tickets = $3

Amount paid by non-students per tickets = $5

So, according to question

S + NS = 771

3 S + 5 NS = 3037

Solve both equations

5 S + 5 NS = 771 × 5

3 S + 5 NS = 3037

so ,

(5 S + 5 NS) - (3 S + 5 NS ) = 3855 - 3037

Or, 2 S = 818

I.e S = $\frac{818}{2}$ = 409

Hence, The number of student tickets were sold = 409 Answer

5 0

3 years ago

The graph of Fx), shown below, resembles the graph of G(x) = x2, but it has

gizmo_the_mogwai [7]

Answer:

F(x) = -3(x + 2)² - 2

Step-by-step explanation:

In the picture attached, the graph is shown.

F(x) has the form a(x - h)² + k, where (h, k) is the vertex of the parabola. We can see in the graph that the vertex is located at (-2, -2), then F(x) = a(x + 2)² - 2. If a > 0 the parabola opens upward, if a < 0 the parabola opens downward. We can see in the graph that the parabola opens downward, then the correct answer is F(x) = -3(x + 2)² - 2

8 0

4 years ago

Carrie's Catering is having a special on their box lunches. The minimum purchase is 13 boxes for $91. Each additional box is $5.

sashaice [31]

There aren't any "following" equations, but I can give you one nonetheless.
y = 91 + 5x
This is because 91 is the minimum and y-intercept, x is each individual box, and the 5 is $5 per box.
Hope this helps!

6 0

3 years ago

The desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5

LekaFEV [45]

Given that t<span>he desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. suppose that the percentage of sio2 in a sample is normally distributed with σ = 0.32 and that $\bar{x}=5.24$ .

</span>
<span>To investigate whether this indicate conclusively that the true average percentage differs from 5.5.

Part A:

From the question, it is claimed that </span><span>t<span>he desired average percentage of sio2 in a certain type of aluminous cement is 5.5</span></span> and we want to test whether the information from the random sample <span>indicate conclusively that the true average percentage differs from 5.5.

Therefore, the null hypothesis and the alternative hypothesis is given by:

$H_0:\mu=5.5 \\ \\ H_a:\mu\neq5.5$

Part B:

The test statistics is given by:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ =\frac{5.25-5.5}{0.32/\sqrt{16}} \\ \\ = \frac{-0.25}{0.32/4} = -\frac{0.25}{0.08} \\ \\ =-3.125$

Part C:

The p-value is given by

$P(z\ \textless \ -3.125)=1-P(z$

Part D:

Because the p-value is less than the significant level α, we reject the null hypothesis and conclude that "</span><span>There is sufficient evidence to conclude that the true average percentage differs from the desired percentage."

Part E:

</span>If the true average percentage is μ = 5.6 and a level α = 0.01 test based on n = 16 is used, what is the probability of detecting this departure from H0? (Round your answer to four decimal places.)

The probability of detecting the departure from $H_0$ is given by

$1-\phi\left(z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right)+\phi\left(-z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right) \\ \\ =1-\phi\left(z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right)+\phi\left(-z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right) \\ \\ =1-\phi\left(z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)+\phi\left(-z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)$

$=1-\phi\left(z_{0.995}+ \frac{-0.1}{0.08} \right)+\phi\left(-z_{0.995}+ \frac{-0.1}{0.08} \right) \\ \\ =1-\phi(2.576-1.25)+\phi(-2.576-1.25) \\ \\ =1-\phi(1.326)+\phi(-3.826) \\ \\ =1-0.90758+0.00007 \\ \\ =0.0925$

Part F:

What value of n is required to satisfy α = 0.01 and β(5.6) = 0.01? (Round your answer up to the next whole number.)

The value of n is required to satisfy α = 0.01 and β(5.6) = 0.01 is given by

$n=\left[ \frac{\sigma(z_{0.005}+z_{0.01})}{\mu_0-\mu} \right]^2 \\ \\ = \left[\frac{0.32(-2.576-2.326)}{5.5-5.6} \right]^2 \\ \\ =\left[\frac{0.32(-4.902)}{-0.1} \right]^2=\left[\frac{-1.56864}{-0.1} \right]^2 \\ \\ =(15.6864)^2=247$

3 0

4 years ago