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2 years ago

Someone pls help plsssss?? ASAP pls

Mathematics

1 answer:

larisa [96]2 years ago

5 0

1)First, understand that whatever your equation is it’s always gonna be less than 100, because she doesn’t wanna get more than 100 ft.
*>100
2) remember that Lainey already dropped 20 feet. The amount of minutes is your unknown variable (x)
3) so, note that the rate is 20 ft/minute. If you multiply 20 by the amount of minutes she has to descend (your unknown variable: x) you cancel out the “minutes” in the 20 ft/minute fraction.
(Ex. 20ft/min *multiply* x [unknown]minutes= 20ftx)
This is what you want since you need all of the terms to be the same unit to solve with algebra.
4) so, to put it all together you have, 20 times x (representing the unknown amount of minutes) plus 20 (representing the amount she has already fallen) is greater than 100

* 20x+20>100

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the answer is 5/6! : )

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3 years ago

Which graph represents the function f(x) = 2x?

Leona [35]

Answer:

This one.

Step-by-step explanation:

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3 years ago

The following data lists the ages of a random selection of actresses when they won an award in the category of Best Actress, al

Valentin [98]

Answer:

a) $p_v =P(t_{(9)}$

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \geq 0$

Alternative hypothesis: $\mu_y -\mu_x$

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679$

The next step is calculate the degrees of freedom given by:

$df=n-1=10-1=9$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v =P(t_{(9)}$

Part b

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".

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

$\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (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: "=-T.INV(0.005,9)".And we see that $t_{\alpha/2}=3.25$

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

$-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469$

$-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069$

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

8 0

3 years ago

An analyst from an energy research institute in California wishes to precisely estimate a 99% confidence interval of the average

Lena [83]

Answer:

190

Step-by-step explanation:

Data provided in the question:

Confidence level = 99%

Therefore,

α = 1% = 0.01

[ from standard normal table ]

z-value for $z_{\frac{\alpha}{2}}= z_{\frac{0.01}{2}}=$ = 2.58

Margin of error, E = $0.06

Standard deviation, σ = $0.32

Now,

n = $(\frac{z_{0.005}\sigma}{E})^2$

Here,

n is the sample size (or the minimum number of gas stations )

on substituting the respective values, we get

= $(\frac{z_{0.005}\sigma}{E})^2$

= $(\frac{2.58\times0.32}{0.06})^2$

= 13.76²

= 189.3376 ≈ 190

Hence,

minimum number of gas stations that she should include in her sample is 190

5 0

3 years ago

Help please and how to did it cuz I’m confused

Brrunno [24]

Answer:

-125

Step-by-step explanation:

Step 1: Find g(-1)

g(x) = -5x²

g(-1) = -5(-1)²

g(-1) = -5(1)

g(-1) = -5

Step 2: Find g(-5)

g(-5) = -5(-5)²

g(-5) = -5(25)

g(-5) = -125

6 0

3 years ago