Explanation Given information: Grass seed A is worth $ 1.40 per pound and seed B is worth of $ 1.75 per pound. And the given table is: SEED A SEED B MIXTUTRE COST OF SEED $1.40 $1.54 AMOUNT (in pounds) x y MIXTUTRE 1.75 y 1.54 ⋅ 50 or Formula used: Use Elimination method to find amount in lb of each Grass seed A and Grass seed B to be mixed in order to make 50 lb of a mixture worth $ 1.54 per pound. The Elimination method for two system of equations are given below: Step1: Write the two system of equations. Step2: Then check whether a same term like 3 y or 3 x are present in the two equations with opposite sign are not. Step3: If the same term present in the two equations with opposite sign then add the two equations by eliminating a variable. Then solve for the other variable. Step4: If the same term is not present in both the equations then multiply a system of equation by an appropriate natural number, so that there are a same term present in both the equations with opposite sign. Then add both system of equations by eliminating a variable. Then solve further for the other variable. Step5: The value of the variable obtained from step 3 or step 4 put in any system of equations and get the value of the eliminated variable. Calculation: Consider Let x = the amount in lb of Grass seed A and y = the amount in lb of Grass seed B. the system of equations can be written as. x + y = 50 …… ( 1 ) 1.40 x + 1.75 y = 77 …… ( 2 ) Use the elimination method. Multiply each side of equation ( 1 ) by − 1.40 and add, the x -terms can be eliminated and solve for y . Multiply equation ( 1 ) by − 1

9th Edition

ISBN: 9780321997173

Author: Marvin L. Bittinger, Judith A. Beecher

Publisher: PEARSON

See similar textbooks

Concept explainers

Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

Videos

Question

Chapter 13.4, Problem 4DE

To determine

To calculate: The amount in lb of each Grass seed A and Grass seed B to be mixed in order to make 50 lb of a mixture worth $1.54 per pound. And complete the provided table:

SEED A SEED B MIXTUTRE

COST OF SEED $1.40 $1.54

AMOUNT (in pounds) x y

MIXTUTRE 1.75y 1.54⋅50 or

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 13.4, Problem 3DE

Chapter 13.4, Problem 5DE

Students have asked these similar questions

~ exp(10). A 3. Claim number per policy is modelled by Poisson(A) with A sample x of N = 100 policies presents an average = 4 claims per policy. (i) Compute an a priory estimate of numbers of claims per policy. [2 Marks] (ii) Determine the posterior distribution of A. Give your argument. [5 Marks] (iii) Compute an a posteriori estimate of numbers of claims per policy. [3 Marks]

How can I prepare for me Unit 3 test in algebra 1? I am in 9th grade.

iid B1 Suppose X1, ..., Xn fx(x), where 2 fx(x) = x exp(−x²/0), 0<< (0 otherwise). (a) Find the maximum likelihood estimator of 0. (b) Show that the MLE is an unbiased estimator of 0. (c) Find the MSE of the MLE. Hint: For parts (b) and (c), you may use integration by parts.

Chapter 13 Solutions

Developmental Mathematics (9th Edition)

Ch. 13.1 - Prob. 1SR Ch. 13.1 - Prob. 2SR Ch. 13.1 - Prob. 1DE Ch. 13.1 - Prob. 2DE Ch. 13.1 - Prob. 3DE Ch. 13.1 - Prob. 4DE Ch. 13.1 - Prob. 5DE Ch. 13.1 - Prob. 6DE Ch. 13.1 - Prob. 7DE Ch. 13.1 - Prob. 8DE

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The amount in lb of each Grass seed A and Grass seed B to be mixed in order to make 50 lb of a mixture worth $ 1.54 per pound. And complete the provided table: SEED A SEED B MIXTUTRE COST OF SEED $1.40 $1.54 AMOUNT (in pounds) x y MIXTUTRE 1.75 y 1.54 ⋅ 50 or

Solution Summary: The author calculates the amount in lb of Grass seeds A and B to be mixed to make a mixture worth 1.54 per pound.

Concept explainers

Videos

To calculate: The amount in lb of each Grass seed A and Grass seed B to be mixed in order to make 50 lb of a mixture worth $1.54 per pound. And complete the provided table:

SEED A SEED B MIXTUTRE

COST OF SEED $1.40 $1.54

AMOUNT (in pounds) x y

MIXTUTRE 1.75y 1.54⋅50 or

Want to see the full answer?

Chapter 13 Solutions

	SEED A	SEED B	MIXTUTRE
COST OF SEED	$1.40		$1.54
AMOUNT (in pounds)	x	y
MIXTUTRE		1.75y	1.54⋅50 or

The amount in lb of each Grass seed A and Grass seed B to be mixed in order to make 50 lb of a mixture worth $ 1.54 per pound. And complete the provided table: SEED A SEED B MIXTUTRE COST OF SEED $1.40 $1.54 AMOUNT (in pounds) x y MIXTUTRE 1.75 y 1.54 ⋅ 50 or

Solution Summary: The author calculates the amount in lb of Grass seeds A and B to be mixed to make a mixture worth 1.54 per pound.

Concept explainers

Videos

To calculate: The amount in lb of each Grass seed A and Grass seed B to be mixed in order to make 50 lb of a mixture worth $1.54 per pound. And complete the provided table: SEED A SEED B MIXTUTRE COST OF SEED $1.40 $1.54 AMOUNT (in pounds) x y MIXTUTRE 1.75y 1.54⋅50 or

Want to see the full answer?

Chapter 13 Solutions

To calculate: The amount in lb of each Grass seed A and Grass seed B to be mixed in order to make 50 lb of a mixture worth $1.54 per pound. And complete the provided table:

SEED A SEED B MIXTUTRE

COST OF SEED $1.40 $1.54

AMOUNT (in pounds) x y

MIXTUTRE 1.75y 1.54⋅50 or