All categories

3 years ago

Find the missing length indicated

Mathematics

2 answers:

faltersainse [42]3 years ago

6 0

Answer:

D) 20

Step-by-step explanation:

Len [333]3 years ago

4 0

Answer:

D) 20

Step-by-step explanation:

they are asking for the longest length of the triangle. as you can see, 20 is just over 15, and from the diagram, the missing length is just over 15 so we can apply our knowledge to come to the answer of 20.

You might be interested in

I need a long essay about may the bridges i burn light the way

aivan3 [116]

Answer:

I will let it lead me down the paths ahead of me. I never in my life thought that Dylan McKay would be a major influence on my growth as a person. But that's the way it seems to be going right now. So wish me luck, and may the bridges I've burned light the way.

this is not that long but I tried

6 0

3 years ago

Each of the 25 students in Mr. McDonald class sold 16 raffle tickets if each ticket cost $15 how much money did Mr. McDonald’s s

Anna [14]

25 * 16 = 400 tickets were sold

400 * $15 = $6000 was raised

7 0

3 years ago

Read 2 more answers

Which statements are true about the trigonometric ratios of this triangle

rosijanka [135]

Answer:

Options (1), (2), (3) and (4)

Step-by-step explanation:

By applying the sine and cosine rules in the given triangle,

sinθ = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

cosθ = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

cos(30°) = $\frac{AC}{AB}$

= $\frac{11\sqrt{3} }{22}$

= $\frac{\sqrt{3}}{2}$

sin(30°) = $\frac{BC}{AB}$

= $\frac{11}{22}$

= $\frac{1}{2}$

cos(60°) = $\frac{BC}{AB}$

= $\frac{11}{22}$

= $\frac{1}{2}$

sin(60°) = = $\frac{AC}{AB}$

= $\frac{11\sqrt{3} }{22}$

= $\frac{\sqrt{3}}{2}$

Options (1), (2), (3) and (4) are the correct options.

3 0

2 years ago

The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,

strojnjashka [21]

Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=400$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=80$ represent the population standard deviation

n=15 represent the sample size

2) Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

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

$400-2.58\frac{80}{\sqrt{15}}=346.708$

$400+2.58\frac{80}{\sqrt{15}}=453.292$

So on this case the 99% confidence interval would be given by (346.708;453.292)

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.98$

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

$400-2.98\frac{80}{\sqrt{15}}=338.445$

$400+2.98\frac{80}{\sqrt{15}}=461.555$

So on this case the 99% confidence interval would be given by (338.445;461.555)

4 0

3 years ago

An engineer is designing a large steel pad to be installed on the deck of an aircraft carrier. Its total volume will be 18 yd^3.

Shtirlitz [24]

Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0. $\overline 1$ yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0. $\overline 1$ yd. = (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel pad, m₁ = ρ × V₁

∴ The mass of the steel pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

∴ C = 4,050 lbs × $55/lb = $222,750

6 0

3 years ago