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2 years ago

Students in a high school mathematics class decided that their term project would be a study of

1 answer:

8 0

The parameter of interest in this scenario is the proportion of all 1,000 students within the school who have "strict" parents or guardians.

<h3>How to calculate a test statistic?</h3>

In Mathematics, the test statistics of a given sample is calculated by using this formula:

$Z=\frac{x\;-\;u}{\frac{\sigma}{\sqrt{n} } }$

<u>Where:</u>

x is the sample mean.

u is the mean.

is the standard deviation.

n is the sample size.

In conclusion, a possible test statistic which can be used to estimate this parameter is the proportion of all students who have "strict" parents or guardians, based on the collected sample.

Read more on test statistic here: brainly.com/question/4621112

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In ∆MNP, m∠N = 90º, NH – altitude, m∠P = 21º, PM = 4 cm. Find MH.

Sveta_85 [38]

Answer:

0.51 cm

Step-by-step explanation:

In right triangle MNP, MP = 4 cm, m∠N = 90°, m∠P = 21°

By the sine definition,

$\sin \angle P=\dfrac{\text{Opposite leg}}{\text{Hypotenuse}}=\dfrac{MN}{MP}\\ \\MN=MP\sin \angle P\\ \\MN=4\sin 21^{\circ}\approx 1.43\ cm$

Now, consider right triangle HMN (it is right because NH is an altitude). By the cosine definition,

$\cos \angle M=\dfrac{\text{Adjacent leg}}{\text{Hypotenuse}}=\dfrac{MH}{MN}\\ \\MH=MN\cos \angle M$

In the right triangle, two acute angles are always complementary, so

$m\angle M=90^{\circ}-m\angle P=90^{\circ}-21^{\circ}=69^{\circ}$

Thus,

$MH=1.43\cos 69^{\circ}\approx 0.51\ cm$

7 0

2 years ago

Martin throws a ball straight up in the air. the equation h(t) = -16t^2 + 40t + 5 gives the height of the ball, in feet, t secon

Lisa [10]

Answer:

2.62 seconds

Step-by-step explanation:

Let

t ----> the time in seconds

h(t) ----> he height of the ball, in feet

we have

$h(t)=-16t^2+40t+5$

we know that

When the ball hits the ground, the height is equal to zero

$-16t^2+40t+5=0$

The formula to solve a quadratic equation of the form

$at^{2} +bt+c=0$

is equal to

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

in this problem we have

$-16t^2+40t+5=0$

$a=-16\\b=40\\c=5$

substitute in the formula

$t=\frac{-40\pm\sqrt{40^{2}-4(-16)(5)}} {2(-16)}$

$t=\frac{-40\pm\sqrt{1,920}} {-32}$

$t=\frac{-40+\sqrt{1,920}} {-32}=-0.12\ sec$

$t=\frac{-40-\sqrt{1,920}} {-32}=2.62\ sec$

therefore

The solution is t=2.62 seconds

4 0

2 years ago

Find tan theta if theta is an angle in standard position and the point with coordinates (4,3) lies on the terminal side of the a

Karolina [17]

Answer: 0.75

Step-by-step explanation:

We can do this by drawing a terminal side in a standard position and then drawing the associated triangle. We then use the Pythagoras theorem to find the missing side the hypotenuse. Then, this will give a Pythagorean Triplet of 3, 4 and 5.

We now have a triangle with the values of:

x = 4

y =3

h^2= 4^2 + 3^2

h^2 = 16 + 9

h^2 = 25

h = 5

Tangent = opposite/adjacent

= y/x

= 3/4

= 0.75

8 0

2 years ago

Kendall had $1000 in a savings account at the beginning of the semester. She has a goal to have at least $350 in the account by

USPshnik [31]

Answer:

11 weeks

Step-by-step explanation:

First we need to check what variables we have.

Beginning Balance = $1000

Goal = $350

Withdrawal = $55 per week

Now let's declare a variable as the number of weeks.

Let x = number of weeks

1000 - 55x = 350

-55x = 350-1000

-55x = -650

Then we divide both sides by -55 to find the value of x.

x = 11.81 or 11 since we're looking for how many weeks in total

Now let's see if we still have 350 if we have a total of 11 as the value of x.

1000 - 55(11) = 350

1000 - 605 = 350

395 = 350

We can see that Kendall will have $395 compared to the $350 goal.

So Kendall can withdraw $55 a week for 11 weeks to still be within her goal of having $350 in her savings account.

6 0

3 years ago

1. In circle C shown below. a tangent has been drawn at point . If angle CBA=28. then explain why the measureof angle DAB must b

Alika [10]

Where’s the picture for the circle C ?

4 0

2 years ago