Answer:
A.0.6
0.6*5yrs=3feet a year
Step-by-step explanation:
Answer:
For 1 miles distance: 7/10 of a mile she runs and 3/10 she walks.
For 4 miles distance (4 times the previous distance) she will run:
4*(7/10) = 18/20 = 2.8 miles
and walk:
4*(3/10) = 12/10 = 1.2 miles.
Answer:
Jewel is 7 years old
John is 1 year old
Dave is 17 years old
Step-by-step explanation:
From the question we are told that:
Jewel age is 
John age 
Dave's age a Year ago

So that Dave.s age this Year could be

Simplifying Equation for Dave's age we have





Dave is 17 years old
Therefore
Johns age is



John is 1 year old
Answer:
A=(16 -5 0 -9)-(3 -4 -11 24)
Step-by-step explanation:
The solution process for any algebraic expression is to "undo" what is done to the variable. Here, matrix A has (3 -4 -11 24) added to it. The next step is to undo that addition, by subtracting that amount from both sides of the equation. The result is ...
A = (16 -5 0 -9)-(3 -4 -11 24)
_____
We assume there are typos in the answer choices listed here.
Answer:
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the probability that a randomly selected individual will be between 185 and 190 pounds?
This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So
X = 190



has a pvalue of 0.8944
X = 185



has a pvalue of 0.7357
0.8944 - 0.7357 = 0.1587
15.87% probability that a randomly selected individual will be between 185 and 190 pounds