Please help me with this very hard question. Determine if each pair of expressions is equivalent. Mark all equivalent expression
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Determine if each pair of expressions is equivalent. Mark all equivalent expressions.
4\left(3m+9-7\right)4(3m+9−7) and 6\left(2m+5-3\right)6(2m+5−3)
12z-3\left(z+5\right)12z−3(z+5) and 19+11z+2\left(-2-z\right)19+11z+2(−2−z)
5x\left(8-4y\right)5x(8−4y) and 2\left(12x-10xy\right)+16x2(12x−10xy)+16x
6-7\left(4k+2\right)6−7(4k+2) and -2\left(k+1\right)−2(k+1)
Determine if each pair of expressions is equivalent. Mark all equivalent expressions.
4\left(3m+9-7\right)4(3m+9−7) and 6\left(2m+5-3\right)6(2m+5−3)
12z-3\left(z+5\right)12z−3(z+5) and 19+11z+2\left(-2-z\right)19+11z+2(−2−z)
5x\left(8-4y\right)5x(8−4y) and 2\left(12x-10xy\right)+16x2(12x−10xy)+16x
6-7\left(4k+2\right)6−7(4k+2) and -2\left(k+1\right)−2(k+1)
1 answer:
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1) ,
,
,
(given)
2) (reflexive property)
3) (SSS)
4) (CPCTC)
5) and
are right angles (perpendicular lines form right angles)
6) and
are right triangles (a triangle with a right angle is a right triangle)
7) (HA)
8) (CPCTC)
Answer:
P-value (t=2.58) = 0.0066.
Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.
As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the population mean μ exceeds 2.4.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the population mean μ exceeds 2.4.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=45.
The sample mean is M=2.5.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the population mean μ exceeds 2.4.