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Sunny_sXe [5.5K]

3 years ago

5

Amber is making a scale drawing of a movie screen. The actual dimensions of a movie screen in her local theater are 60 ft wide a

nd 20 ft tall. In her scale drawing, the screen has a width of 9 in. What will the height of the screen on her drawing be?

1 answer:

Annette [7]3 years ago

7 0

Answer:

3 inches

60x = 9 x 20

60x = 180

x = 3

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If cos B is 4/5 what is sin a?

Keith_Richards [23]

Answer:

sin(A+B)=sinAcosB+cosAsinB. Putting the values in. sin(A+B)=(4/5)(−3/5)+(3/5)(−4/5). sin(A+B)=−24/25.

8 0

3 years ago

A personnel director in a particular state claims that the mean annual income is the same in one of the state's counties (Coun

Shalnov [3]

Answer:

a) The hypothesis that the mean annual income is the same could not be rejected. There is no enough evidence to claim they are different.

b) The critical values for this two-sided test are t=±1.711.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>(a) Identify the claim and state H0 and Ha. </em>

<em> Which is the correct claim below?</em>

<em>(b) Find the critical value(s) and identify the rejection region(s). </em>

<em> Enter the critical value(s) below.</em>

We have an hypothesis test on the difference of two means.

The null and alternative hypothesis are:

$H_0: \mu_a-\mu_b=0\\\\H_a: \mu_a-\mu_b\neq 0$

The claim of the alternative hypothesis is that the mean annual income is different in County A and County B.

The null hypothesis is that the mean annual income is equal in both counties.

The significance level is α=0.10.

The sample from County A has a mean of S40,400, a s.d. of $8,700 and a sample size of 18 residents.

The sample from County B has a mean of S39,200, a s.d. of $6,000 and a sample size of 8 residents.

The standard error of the difference of means is:

$\sigma_d=\sqrt{\frac{\sigma_a^2}{n_a}+\frac{\sigma_b^2}{n_b}}=\sqrt{\frac{8700^2}{18}+\frac{6000^2}{8}}=\sqrt{8705000}=2950$

The degrees of freedom are:

$df=n_a+n_b-2=18+8-2=24$

Then, the test statistic is:

$t=\frac{\Delta M-\Delta \mu}{\sigma_d} =\frac{(40,400-39,200)-0}{2950} =\frac{1200}{2950}=0.407$

For a statistic t=0.407, and df=24, the P-value is P=0.69. As the P-value is bigger than the significance level, the null hypothesis failed to be rejected.

If we would use the critial value approach, we would have to calculate the critical values for t, for df=24, two sided test and α=0.10.

The critical values, looking in a table, are t=1.711.

5 0

3 years ago

A students cost for last semester was $2100. She spent $420 of that on books. What percent of last semesters college cost was sp

kati45 [8]

the answer would be 20% on the books

3 0

3 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$

$f'(x)=12x^2+26x-12$

$f''(x)=24x+26$

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here: x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

$24(-2.6)+25$

$-37.5$

So we have a maximum at x=-2.6.

$24(.4)+25$

$9.6+25$

$34.6$

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

$y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56$

I'm shoving this into a calculator:

$y=-58.4654411$

So the approximate relative minimum is (0.4,-58.5).

If you graph $y=4x^3+13x^2-12x-56$ you should see the graph taking a dip at this point.

3 0

3 years ago

Two bags contain blue and red marbles. The first bag contains 3 blue and 5 red marbles. The second bag contains

dimaraw [331]

Answer: 1/8

<u>Step-by-step explanation:</u>

First Bag and Second Bag

$\dfrac{3\ blue}{8\ total}$ x $\dfrac{2\ blue}{6\ total}$ = $\dfrac{6}{48}$

which reduces to $\boxed{\dfrac{1}{8}}$

8 0

3 years ago