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2 years ago

Please help!

Mathematics

1 answer:

Nimfa-mama [501]2 years ago

7 0

Answer:

As "x" increases "y" *decreases*

y as a function of x is not *constant*

Therefore the function is *nonlinear*

For all values of x, the function value y *≥* 0.

The y intercept of the graph is *8*

x=1 function value of y is *5*

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THIS IS MY LAST ONE PLEASE HELP

tia_tia [17]

Answer:

x=56

Step-by-step explanation:

okay so

x=180- (64+60)

x= 180-124

x= 56

4 0

2 years ago

Finding surface area if an triangular prism

SVEN [57.7K]

1)find the area of a triangle. Multiply it by 2 or 4 depending on if they are the same size
2)find the area of the base
3)find the area of the other two sides if you haven’t already.
4) add them all together

8 0

3 years ago

An airplane flies due north from ikeja airport for 500km.It then flies on a bearing of 060 from a further distance of 300km befo

Nata [24]

Answer:

482 km

63.94 degrees

Step-by-step explanation:

to solve this question we will use the cosine rule. For starters, draw your diagram. From point A, up north is 500km and 060 from there, another 300. If you join the point from the road junction back to the starting point, yoou have a triangle.

Cosine rule states that

C = $\sqrt{A^{2} + B^{2} -2AB cos(c) }$

where both A and B are the given distances, 500 and 300 respectively, C is the 3rd distance we're looking for and c is the given angle, 060

solving now, we have

C = $\sqrt{500^{2} + 300^{2} -2 * 500 * 300 cos(60) }$

C = $\sqrt{250000 + 90000 - [215000 cos(60) }]$

C = $\sqrt{340000 - [215000 * 0.5 }]$

C = $\sqrt{340000 - [107500 }]$

C = $\sqrt{232500}$

C = 482 km

The bearing can be gotten by using the Sine Rule.

$\frac{sina}{A}$ = $\frac{sinc}{C}$

sina/500 = sin60/482

482 sina = 500 sin60

sina = $\frac{500 sin60}{482}$

sina = 0.8983

a = sin^-1(0.8983)

a = 63.94 degrees

6 0

2 years ago

Write the place vslue of 7in 67,689

Tems11 [23]

The value of 7 can be wrote as

7,000 or seven thousand

8 0

3 years ago

The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime

Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

$Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334$

X[bar] ± $Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }$

174.5 ± $2.334* \frac{6.9}{\sqrt{50} }$

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0

2 years ago