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2 years ago

9.87 multiplied by 19

Mathematics

1 answer:

klasskru [66]2 years ago

5 0

9.87 x 19 = 187.53

mark brainliest if possible :)

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Is this right, can someone please explain so?

Orlov [11]

You have to turn them in to the lowest common denominator

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3 years ago

Need help ASAP!!

Ilia_Sergeevich [38]

An object that is in motion as a projectile follows a path or trajectory of a parabola

The function and values are;

a) The equation of the quadratic function is; $\underline{y = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}$

b) The maximum height of the ball is approximately 7.334 m

c) Horizontal distance at maximum height 18.8 meters

Reason:

a) Known parameters are;

Let f(x) = a·x² + b·x + c represent the equation of the parabola modelling the path of the ball, we have;

Points on the path of the parabola = (0, 0), (35, 1.5), 37, 0)

Plugging the values gives;

0 = a·0² + b·0 + c

Therefore, c = 0

1.5 = 35²·a + 35·b

0 = 37²·a + 37·b

Solving gives;

a = -3/140, b = 111/140

The equation of the quadratic function is therefore;

$\underline{y = f(x) = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}$

b) The maximum height is given by the vertex of the parabola

The x-coordinate at the vertex is the point $-\dfrac{b}{2 \cdot a}$

Which gives;

$x-coordinate = \dfrac{\frac{111}{140} }{2 \times \dfrac{3}{140} } = 17.5$

The maximum height is therefore;

$f(x)_{max} = \dfrac{111}{140} \times 17.5 - \dfrac{3}{140} \cdot 17.5^2 \approx 7.334$

The maximum height of the ball is approximately 7.334 m

c) The distance the ball has travelled to horizontally is given by half of the range, R as follows;

The range of the motion, R = 37 meters

$Horizontal \ distance \ to \ maximum \ height = \dfrac{R}{2}$

Therefore;

$Horizontal \ distance \ to \ maximum \ height = \dfrac{37}{2} = 18.5$

The distance the ball has travelled horizontally to reach the maximum height horizontally 18.5 meters

Learn more about the trajectory of a projectile here:

brainly.com/question/13646224

8 0

3 years ago

A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

The standard error is given by the following formula:

$SE=\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$

And replacing we have:

$SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

Part c: What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

Confidence interval

Now we have everything in order to replace into formula (1):

$3.92-1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=2.009$

$3.92+1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=5.830$

So on this case the 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

7 0

3 years ago

A new car is purchased for 23700 dollars. The value of the car depreciates at

Tpy6a [65]

Answer:

5 years (about)

Step-by-step explanation:

$23,700 X .13 = $3,081

$3,081 X 5 = $15,405

$23,700 - $15,405 = $8,295

Good luck! hope this helps.

(If you've noticed a mistake in my logic, please comment any corrections)

7 0

3 years ago

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astra-53 [7]

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3 years ago

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