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2 years ago

12

Karen wants to rent a car for 5 days, and shes interested in the loss damage waiver but not the collision damage waiver. If the

daily rate is $35 and the LDW is $55, what is her total cost?
A.$450
B.$238
C.$230
D.$228

2 answers:

DanielleElmas [232]2 years ago

8 0

Add $35 + $55 and you get 90, multiply 90 by 5 and you get $450. so the answer is A. hope this helped.

vlabodo [156]2 years ago

7 0

The answer is 230 hope this helps

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2 years ago

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A shed has dimensions of 12m in length and 5 m in width. Both the length and width are increased by the same amount in order to

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The amount by which the length and width of a shed can be increased to

more than double the area, depends on the initial dimensions.

The amount that can be added to both the length and the width to increase the floor area by more than double the original area is more than <u>3 meters</u>.

Reasons:

The given parameters of the shed are;

The length of the shed = 12 m

Width of the shed = 5 m

The amount by which the length and the width are increased = The same amount

The new area after the increase in the length and width of the shed = More than double the initial area

Required:

The amount of increase in the length.

Solution:

Let the amount by which the length and width are increased = x

We have;

Initial area of the shed = 12 m × 5 m = 60 m²

The new area = (12 + x) × (5 + x) > 2 × 60

By multiplication, we get;

(12 + x) × (5 + x) = x² + 17·x + 60 > 2 × 60 = 120

x² + 17·x + 60 - 120 > 120 - 120 = 0

x² + 17·x - 60 > 0

By factorization, we get;

(x + 20)·(x - 3) > 0

x > -20, or x > 3

The increase (positive) amount of the solution is x > 3

Therefore, <u>the amount by which both the length and the width can be increased to more than double the area is x > 3 meters</u>

Learn more about the area of a rectangle here:

brainly.com/question/16410706

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2 years ago

A coin is tossed 71 times and 34 heads are observed. Is the coin fair? Use a 98% level confidence interval to base your inferenc

Fofino [41]

Answer:

a) $\hat p=\frac{34}{71}=0.48$ estimated proportion of heads observed.

b) $Se= \sqrt{\frac{0.479(1-0.479)}{71}}=0.0593$

c) $z_{\alpha/2}=2.33$

d) $0.479 - 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.341$

e) $0.479 + 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.617$

f) In order to be fair the coin needs to have included the 0.5 on the interval, and for this case it's included, so we can say that at 2% of significance we can say that the coin is fair.

Step-by-step explanation:

Notation and definitions

$X=34$ number of heads observed.

$n=71$ random sample taken

$\hat p=\frac{34}{71}=0.479$ estimated proportion of heads observed.

$p$ true population proportion of heads observed.

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 population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The point estimator for the population proportion of heads is: Answer with two decimal precision.

$\hat p=\frac{34}{71}=0.48$ estimated proportion of heads observed.

The standard error in this estimate is: Answer with four decimal precision.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And the standard error is given by:

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

$Se= \sqrt{\frac{0.479(1-0.479)}{71}}=0.0593$

The correct table value for the 98% level confidence interval is: Answer with two decimal precision

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

The lower end point of the CI is: Answer with three decimal precision.

And replacing into the confidence interval formula we got:

$0.479 - 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.341$

The upper end point of the CI is: Answer with three decimal precision.

$0.479 + 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.617$

And the 90% confidence interval would be given (0.341;0.617).

Based on the confidence interval, it is plausible that the coin is fair.

In order to be fair the coin needs to have included the 0.5 on the interval, and for this case it's included, so we can say that at 2% of significance we can say that the coin is fair.

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3 years ago

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Answer:

$27,72

Step-by-step explanation:

The information that we have its

The cost gas its $3.60 per gallon
Irene's car had 5.3 gallons
Irene's car tank is 13 gallons

so now for the know, How much did Irene spend on gas?

First

The gas that its necesary for filled de tank its

13gallons-5,3gallons=7,7 gallons

The cost for 7,7 its

7,7g*3,6$/g= $27,72

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3 years ago

Last Wednesday, students could choose ham or turkey sandwiches for lunch. The cafeteria made 50 sandwiches in all, 5 of which we

PSYCHO15rus [73]

Answer:

90%

Step-by-step explanation:

50%×2 100%

5%×2 10%

45%×2 90%

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3 years ago