The value of x in the equation is 2.
<h3>What is indices?</h3>Indices singular index is a branch of algebra that deals with the power or exponent of variables.
Laws of indices help us to evaluate indicial expressions and equations.
Analysis:
= 256
Here 4 is the base number while 8-2x is the exponent or index.
So for us to evaluate this, we need to write 256 in its index form having its base as 4 also.
so 256 in base 4 index form is
=
since both sides have the same base number, we equate only their exponent
8-2x = 4
-2x = 4 - 8
-2x = -4
x = -4/-2 = 2
Learn more about indices: brainly.com/question/10339517
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Step-by-step explanation:
3(p2q + pq2 - pq) - 6pq + 3pq2 =
3[-20 × 2 + (-20) × 2 - (-20)] - 6 × (-20) + 3 × 2 × (-20) =
3[-40 + (-40) + 20] - 6 × (-20) + 3 × 2 × (-20) =
3(-40 - 40 + 20) - 6 × (-20) + 3 × 2 × (-20) =
3(-80 + 20) - 6 × (-20) + 3 × 2 × (-20) =
3 × (-60) - 6 × (-20) + 3 × 2 × (-20) =
-180 + 120 + 6 × (-20) =
-180 + 120 + (-120) =
-180 + 120 - 120 =
-60 - 120 = -180
1190 pairs of shoes are defective
<em><u>Solution:</u></em>
Given that shoe company found that 4.25% of the production is defective
The company made 28,000 pairs of shoes
To find: Number of pairs of defective shoes
From given information, 4.25% of the production is defective which means 4.25% of 28,000 pairs of shoes
Defective shoes = 4.25% of 28,000
Defective shoes
Defective shoes
Thus 1190 pairs of shoes were defective
Answer:
(a) We reject our null hypothesis.
(b) We fail to reject our null hypothesis.
(c) We fail to reject our null hypothesis.
Step-by-step explanation:
We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.
A random sample of 18 pens is selected.
<u><em>Let </em></u><u><em> = true average writing lifetime under controlled conditions</em></u>
So, Null Hypothesis, :
10 hr {means that the true average writing lifetime under controlled conditions is at least 10 hr}
Alternate Hypothesis, :
< 10 hr {means that the true average writing lifetime under controlled conditions is less than 10 hr}
<u>The test statistics that is used here is one-sample t test statistics;</u>
T.S. = ~
where, = sample mean
s = sample standard deviation
n = sample size of pens = 18
n - 1 = degree of freedom = 18 -1 = 17
<u>Now, the decision rule based on the critical value of t is given by;</u>
(a) Here, test statistics, t = -2.4 and level of significance is 0.05.
<em>Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.</em>
Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.
(b) Here, test statistics, t = -1.83 and level of significance is 0.01.
<em>Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.</em>
Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.
(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.
<em>Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.</em>
Here, clearly the value of test statistics is more than the critical value of t as -1.74 < 0.57, so we fail to reject our null hypothesis.