All categories

2 years ago

the sides of a triangle are in the ratio 3:4:5 what is the length of each side if the perimeter of the triangle is 90cm

1 answer:

6 0

P of triangle = a + b = c = 90

<span>ratio 3:4:5
so a = 3x, b = 4x , c = 5x

3x + 4x + 5x = 90
12x = 90
x = 90/12 = 7.5

a = 3x = 3 * 7.5 = 22.5
b = 4x = 4 * 7.5 = 30
c = 5x = 5 * 7.5 = 37.5

Answer:
1st side = 22.5
2nd side = 30
3rd side = 37.5

Double check: 22.5 + 30 + 37.5 = 90 (perimeter)</span>

You might be interested in

A tobacco company claims that its best-selling cigarettes contain at most 40 mg of nicotine. The average nicotine content from a

tatuchka [14]

Answer:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

$df=n-1=15-1=14$

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

Step-by-step explanation:

Information provided

$\bar X=42.6$ represent the average nicotine content

$s=3.7$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =40$ represent the value to check

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to verify if the nicotine content of the cigarettes exceeds 40 mg , the system of hypothesis are:

Null hypothesis: $\mu \leq 40$

Alternative hypothesis: $\mu > 40$

The statistic for this case would be:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replcing we got:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

The degrees of freedom are:

$df=n-1=15-1=14$

The p value would be:

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

6 0

3 years ago

Which is an example of closed-end credit?

kumpel [21]

the answer is C) home loan

5 0

2 years ago

Read 2 more answers

Find the arc length of a central angle 300 degrees in a circle whose radius is 2 inches? help now ASP I NEED HELP A.10 π/ 3 in

NeTakaya

$\text{Arc length } = \frac{\theta}{360}2\pi r$

$= (\frac{300}{360})2\pi (2)$

$= 4\pi(\frac{300}{360})$

$= 4\pi(\frac{5}{6})$

$= \frac{20\pi}{6}$

$= \frac{10\pi}{3} \text{ in}$

5 0

3 years ago

Read 2 more answers

Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240<br><br> dx/dt=

katrin2010 [14]

Answer:

The value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

Step-by-step explanation:

The given equation is

$x^4=8y^5-240$

We need to find the value of $\frac{dx}{dt}$ .

Differentiate with respect to t.

$4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0$ $[\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]$

$4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}$

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

$4x^3\frac{dx}{dt}=40(2)^4(1)$

$4x^3\frac{dx}{dt}=40(16)(1)$

$4x^3\frac{dx}{dt}=640$

Divide both sides by 4x³.

$\frac{dx}{dt}=\frac{640}{4x^3}$

$\frac{dx}{dt}=\frac{160}{x^3}$

Therefore the value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

4 0

2 years ago

which of th following are characteristics of the reciprocal parent functiona. it has a horizontal asymptote at y=0b. it passes t

Tcecarenko [31]

All of the optiona are characteristics of the reciprocal parent function except option b. (i.e. the reciprocal parent function do not pass through the point (1, 0)

8 0

3 years ago