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4 years ago

Please help and thank you.

Mathematics

1 answer:

Marina CMI [18]4 years ago

8 0

Answer:

400 ft elevation at 1 km and 3 km from the start
300 ft/km average rate of change over the first 4 km

length = 3.5 cm
width = 3.5 cm
height = 7.0 cm

6 mo: $1018.14
5 yrs: $1196.89
avg incr: $3.28 per month

Step-by-step explanation:

A graphing calculator is handy for solving problems involving cubic polynomials. You're interested in where the elevation is 400 ft. Since the value of f(x) is in hundreds of feet, you want to find x such that f(x) = 4.

Values of x where that is the case are x=1 and x=3, representing distances of 1 km and 3 km from the start of the road.

The average rate of change of elevation is the difference in elevation divided by the difference in distance from the start:

average rate of change = (f(4) -f(0) hundred ft)/((4 - 0) km)

= (19 -7)/4 hundred feet/km

= 12/4 hundred feet/km = 300 ft/km

___

You want to find x when ...

A(x) = 122.5 cm²

10x² = 122.5 cm² . . . . substitute the given expression for A(x)

x² = 12.25 cm² . . . . . . divide by 10

x = √(12.25 cm²) = 3.5 cm . . . . take the square root

The diagram tells you ...

length = width = x = 3.5 cm

height = 2x = 7.0 cm

___

Evaluate the given expression for the different values of m:

$1000·1.003^6 ≈ $1018.14 . . . . 6-month value

$1000·1.003^60 ≈ $1196.89 . . . . 5-year value

The increase is $1196.89 -1000.00 = $196.89. That increase took place over 60 months, so the average increase per month is ...

$196.89/(60 mo) ≈ $3.28 per mo . . . . average per month over 5 years

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3 years ago

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Step-by-step explanation:

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2 years ago

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3 years ago

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I already got the answers for these but I wanted to check them. I’m supposed to find the area and use 3.14 for pi. Thanks.

RideAnS [48]

<u>Answer:</u>

10.

A = 12² - 2(6×5)

A = 144 - 2×30

A = 144 - 60

A = 84 in²

11.

A = π(12/2)² - (π3² + 3×6)

A = π36 - (π9 + 18)

A = π36 - (π3² + 18)

A = 113.04 - (28.26 + 18)

A = 66.8 yd²

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3 years ago

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of

kolezko [41]

Answer:

a) Parameter of interest $p$ representing the true proportion of the plates have blistered.

b) Null hypothesis: $p\leq 0.1$

Alternative hypothesis: $p > 0.1$

c) $z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33$

d) For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:

$z_{critc}= 1.28$

e) For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1

f) $p_v =P(z>1.33)=0.0917$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

Part a

Parameter of interest $p$ representing the true proportion of the plates have blistered.

X=14 represent the number of the plates have blistered.

$\hat p=\frac{14}{100}=0.14$ estimated proportion of the plates have blistered.

$p_o=0.1$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Part b: Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that more than 10% of all plates blister under such circumstances.:

Null hypothesis: $p\leq 0.1$

Alternative hypothesis: $p > 0.1$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Part c: Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33$

Part d: Rejection region

For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:

$z_{critc}= 1.28$

Part e

For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1

Part f

Since is a right taild test the p value would be:

$p_v =P(z>1.33)=0.0917$

7 0

3 years ago

Please help and thank you. ​

Please help and thank you.