<span>pHProblem 4- Determine the hydrogen concentration of bloodTutorial to help you with this calculation.<span>4.What is the approximate hydrogen ion concentration of blood with pH 7.4?</span><span> <span><span>TutorialIn this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,− pH = log [H+] ,[H+] = 10−pH,by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M. </span></span></span></span>
Answer:
Step-by-step explanation:
If k is the number of classes and n is the number of observations, then for number of classes we should select the smallest k such that 2^k > n.
<u>We have n = 50 and:</u>
- 2^5 = 32 < 50
- 2^6 = 64 > 50
As per above described 2 to k rule, we are taking k = 6.
So 6 classes should be used.
Answer: the cost of one package of macadamia nut chip cookie dough is $1.5
the cost of one package of triple chocolate cookie dough is $2.5
Step-by-step explanation:
Let x represent the cost of one package of macadamia nut chip cookie dough.
Let y represent the cost of one package of triple chocolate cookie dough.
Mrs. Julien’s class sold 25 packages of macadamia nut chip cookie dough and 30 packages of triple chocolate cookie dough for a total of $112.50. This means that
25x + 30y = 112.5 - - - - - - - - - - - - -1
Mrs. Castillejo’s class sold 8 packages of macadamia nut chip cookie dough and 45 packages of triple chocolate cookie dough for a total of $124.50. This means that
8x + 45y = 124.5 - - - - - - - - - - -2
Multiplying equation 1 by 8 and equation 2 by 25, it becomes
200x + 240y = 900
200x + 1125 = 3112.5
Subtracting, it becomes
- 885y = - 2212.5
y = - 2212.5/- 885
y = 2.5
Substituting y = 2.5 into equation 1, it becomes
25x + 30 × 2.5 = 112.5
25x + 75 = 112.5
25x = 112.5 - 75 = 37.5
x = 37.5/25 = 1.5
The method used is the elimination method. It is more convenient to use.
Given:
Segment MN has endpoints at M(-6, -3) and N(9,7).
Point Q lies on MN such that MQ:QN = 3:2.
To find:
The coordinates of point Q.
Solution:
Section formula: If a point divides of line segment whose end points are 
and 
in m:n, then the coordinates of that points are:
Segment MN has endpoints at M(-6, -3) and N(9,7) and Point Q lies on MN such that MQ:QN = 3:2. By using section formula, we get
Therefore, the coordinates of point Q are (3,3).
