Nformation about the students at jackson high school is shown in the table.

a 4-column table has 4 rows. the first column has entries tenth grade, eleventh grade, twelfth grade, total. the second column is labeled girls with entries 194, blank, 156, 523. the third column is labeled boys with entries blank, 179, blank, 511. the fourth column is labeled total with entries 376, blank, 306, 1,034.

which statement must be true?

out of the 352 eleventh graders at the school, 173 of them are girls.
out of the 511 boys at the school, 150 of them are tenth graders.
out of the 523 tenth graders at the school, 173 of them are boys.
out of the 376 girls at the school, 182 of them are eleventh graders

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.

step-by-step explanation:

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.

$655.61 which is option c

step-by-step explanation:

price of house =   $146,000

initial rate = 3.5%

mortgaged paid in this = 3.5 % of $146,000

                                        =

                                        =$5110

the total years would be = 7

total mortgaged paid = 7 * $5110

                                      =$35770

remaining money = $146000-35770

                                =$110230

new mortgaged payment = 5 % of 110230

                                            =  

                                            =$655.61

graph b

step-by-step explanation:

one solution- graph a (clear intercepting point.)

no solution- graph b (they're parallel, there will never be a point of interception.)

infinite solutions- graph c (all points intercept.)

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages