Aconservationist wants to know if the average water level in horseshoe lake is more than the average water level in swan lake.
1. a sample of 23 observations from horseshoe lake has a mean of 43 meters and a standard deviation of 3.2 meters.
2. a sample of 23 observations from swan lake has a mean of 38 meters and a standard deviation of 2.4 meters. test his hypothesis at α = 0.01.
group of answer choices:
o reject h0; it appears that the average water level in horseshoe lake is more than the average water level in swan lake.
o fail to reject h0; it appears that the average water level in horseshoe lake is not more than the average water level in swan lake.
o not sufficient information to decide

The question is incomplete:

Lisa and Tomas Lopez are buying a new minivan that has a base price of $22,510.the options total $8,062 and destination charge is $770. The dealer's cost is estimated to be 91% of the base price and 88% of the options.

-What is the sticker price of the vehicle?

-Find the dealer's percent of the base price.

-Find the dealer's percent of the options price.

-What is the estimated dealer's cost?

-What is the sticker price of the vehicle? $31,342

-Find the dealer's percent of the base price: $20,484

-Find the dealer's percent of the options price: $7,094

-What is the estimated dealer's cost?  $28,348

Step-by-step explanation:

-The sticker price is the result of adding the base price plus the options plus the destination charge.

Sticker price= base price + options + destination charge

Sticker price= $22,510+$8,062+$770

Sticker price= $31,342

-Dealer's percent of the base price: It is  91% of the base price:

$22,510*91%= $20,484

-Dealer's percent of the options price: It is  88% of the options:

$8,062*88%= $7,094

-Estimated dealer's cost= Percent of base price+percent of options price+destination charge

Estimated dealer's cost= $20,484+$7,094+$770

Estimated dealer's cost=$28,348