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2 years ago

An identity thief watches someone enter a 4-digit PIN and manages to see the first two digits. If the thief tries guessing

Mathematics

1 answer:

Otrada [13]2 years ago

3 0

Answer:

100

Step-by-step explanation:

i dont know if im right though i hope this helps ^^

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A new car that sells for $28,000 depreicates 25% each year. Write a function that models the value of the car. Find the value of

BartSMP [9]

Answer:

Step-by-step explanation:

using the information on the problem a function can can be made

f(n)= 28000(0.75)×

where x is the amount of years

plug in 4 for x in the equation to get

f(4)=8859.375

5 0

2 years ago

Mary Ann needs 3/7 cup of sugar to bake a cake she has 3/5 cup does she have enough to bake the cake yes or no

IrinaVladis [17]

Answer:

Mary Ann does have enough sugar

Step-by-step explanation:

3/5 is greater than 3/7 which means that she does have enough sugar. This is because the denominator 5 is smaller than the denominator 7, and since they have the same numerator, the fraction with the smaller denominator is greater.

5 0

2 years ago

Read 2 more answers

How do you do a question like this? 80% as a fraction in simplest form.

Gemiola [76]

Take away 0.20 of the amount.

3 0

3 years ago

On the trip from KC to Springfield, the church van consumed 15 gallons of gas to go 179 miles. If gas cost $2.58 per gallon, wha

Otrada [13]

Answer:

$77.4

Step-by-step explanation:

GIVEN: On the trip from KC to Springfield, the church van consumed $15$ gallons of gas to go $179$ miles. If gas cost $\$2.58$ per gallon.

TO FIND: Cost of a round trip to Springfield.

SOLUTION:

Distance of springfield from KC $=179\text{ miles}$

As it is a round trip total distance $=2\times179=358\text{ miles}$

As $15$ gallons gas used for $179$ miles distance

gas required for $1$ mile $=\frac{15}{179}$ gallons

gas required for $358$ miles $=\frac{15}{179}\times2=30$ gallons

cost of one gallon of gas $=\$2.58$

cost of $30$ gallons of gas $=30\times2.58=\$77.4$

Hence cost of roundtrip to Springfield is $77.4

6 0

2 years ago

When fishing off the shores of Florida, a spotted seatrout must be between 10 and 30 inches long before it can be kept; otherwis

Mademuasel [1]

Answer:

There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so $\mu = 22, \sigma = 4$ .