Let
x-------> the first number
y-------> the second number
P-----> product
we know that
x+2y=56----------> 2y=56-x-----> y=28-0.5x------> equation 1
P=x*y----> equation 2
substitute equation 1 in equation 2
P=x*[28-0.5x]-----> P=28x-0.5x²
using a graph tool
see the attached figure
the vertex is the point (28,392)
that means
for x=28
the product is 392 (maximum)
find the value of y
y=28-0.5x----> y=28-0.5*28----> y=14
the answer isthe numbers are 28 and 14
Answer:
-4
Step-by-step explanation:
Since the total score <em>I</em><em>n</em><em>c</em><em>r</em><em>e</em><em>a</em><em>s</em><em>e</em><em>d</em><em> </em>by eight than the two matching cards must have been -4.
(total of four cards) + (two matching cards) = total score
(total of four cards) + (two cards × -4) = total score
(total of four cards) - 8 = total score
* take away two matching cards by adding 8 on both sides*
(total of four cards) = total score + 8
12 because if u multiply 3*4 it is 12 and 4*3 is 12 so they both have 12 in common.
Step-by-step explanation:
heght x base
13 x 18
the hight can be measured by the paralel
Answer:
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
Step-by-step explanation:
We formulate null and alternate hypotheses are
H0 : u1 < u2 against Ha: u1 ≥ u 2
Where u1 is the group tested after they were awake for 24 hours.
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≥ t (0.05, 13) = 1.77
Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13
Here the difference between the sample means is x`1- x`2= 35-24= 11
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/13 [ 120²+360²]
Sp = 105.25
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= 11/ 105.25 √1/5+ 1/10
t= 11/57.65
t= 0.1908
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
