alis dog weighs 8 times as much as her cat. together, the two pets weigh 54 pynds. how much does Alis dog weigh?
1 answer:
The weight of the Alis dog is pounds.
Further explanation:
Consider the weight of the Alis dog as pounds and weight of the cat as
pounds.
Since, the weight of the Alis dog is 8 times the weight of the cat. It can be written as follows:
......(1)
Now, the total weight of the cat and dog are given as 54 pounds. It can be written as follows:
......(2)
From equation (1), substitute for
in equation (2).
Thus, the weight of the cat is 6 pounds.
Substitute 6 for in equation (1) to obtain the weight of Alis dog as follows:
Thus, the weight of the Alis dog is 48 pounds.
Learn more:
1. Which function has an inverse that is also a function? {(–1, –2), (0, 4), (1, 3), (5, 14), (7, 4)} {(–1, 2), (0, 4), (1, 5), (5, 4), (7, 2)} {(–1, 3), (0, 4), (1, 14), (5, 6), (7, 2)} {(–1, 4), (0, 4), (1, 2), (5, 3), (7, 1)}
2. A given line has the equation 10x + 2y = −2. what is the equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12)? y = ( )x + 12
3. What are the domain and range of the function f(x) = 3x + 5?
Answer Details :
Grade: Middle School .
Subject: Mathematics .
Chapter: linear equation.
Keywords:
lines, equation, pound, Alis dog, cat, weight, linear equation, zeros, x,y,x=8y,x+y=54, function, substitution, pets, direct substitution, elimination, graph, middle term factorization, fraction.
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Answer:
The null and alternative hypothesis are:
At a significance level of 0.05, there is enough evidence to support the claim that the population percentage of SiO2 is signficantly different from 5.5%.
P-value = 0.000004.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the population percentage of SiO2 is signficantly different from 5.5%.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=16.
The sample mean is M=5.21.
The standard deviation of the population is known and has a value of σ=0.26.
We can calculate the standard error as:
Then, we can calculate the z-statistic as:
This test is a left-tailed test, so the P-value for this test is calculated as:
As the P-value (0.000004) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population percentage of SiO2 is signficantly different from 5.5%.