All categories

3 years ago

Use this numbers and make it into an expression.

Mathematics

1 answer:

maks197457 [2]3 years ago

8 0

4(25 x 3) +15 to the tenth power
Is this what u we’re looking for? Hope this helps!

You might be interested in

Ellen is opening. Cookie shop eat a local middle school . She randomly surgery’s students to determine The types of cookies they

saw5 [17]

Complete Question:

Ellen is opening. Cookie shop eat a local middle school . She randomly surveys students to determine The types of cookies they would buy at the cookie shop the results of the surveys are below based on the survey. Which statement is not true?

A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

B) 22½% of the students surveyed would buy a sugar cookie.

C) If 240 students were to purchase a cookie from the store, approximately 66 students will purchase a peanut butter cookie.

D) Half of the students prefer chocolate chips or oatmeal raisin cookies.

(Table showing the results of her survey is in the attachment below)

Answer:

A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

Step-by-step Explanation:

STEP 1:

In order to ascertain which of the statements given in the options that is NOT TRUE, let's express the given number of students that would buy the different types of cookies in their PROPORTIONS AND PERCENTAGES in the survey result.

Thus:

Type of Cookie==> no of students => Proportion (no of students of a cookie type ÷ total no of students) => % (each proportion for a cookie type)

Chocolate Chip==> 25 => 0.3125 (25÷80) => 31.25% (0.3125 × 100)

Oatmeal Raisin==> 15 => 0.1875 (15÷850) => 18.75% (0.1875 × 100)

Peanut Butter== 22 => 0.275 (22÷80) => 27.5% (0.275 × 100)

Sugar ==> 18 => 0.225 (18÷80) => 22.5% (0.225 × 100)

STEP 2:

Next step is to consider each statement given in the question to see if they are true or not.

==>OPTION A: If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

To find out if this is true, multiply the proportion of students who would be buy a sugar cookie (0.255) by 160 = 0.255 × 160 = 36.

If 160 buy a cookie, approximately 36 students would buy a sugar.

Option A IS NOT TRUE.

OPTION B: 22½% of the students surveyed would buy a sugar cookie.

From our calculation in STEP 2, we have:

Sugar ==> 18 => 0.225 (18÷80) => 22.5% (0.225 × 100).

Option B is TRUE. 22.5% of the students would go for sugar cookie.

OPTION C: If 240 students were to purchase a cookie from the store, approximately 66 students will purchase a peanut butter cookie.

Number of students that would buy the peanut butter if 240 students were to get a cookie = proportion of students that opt for peanut butter cookie in the survey (0.275) × 240 = 0.275 × 240 = 66.

Option C is TRUE.

OPTION D: D) Half of the students prefer chocolate chips or oatmeal raisin cookies.

Proportion of students who prefer chocolate chips or oatmeal raisin cookies = 0.3125 + 0.1785 = 0.491

This is approximately 0.5 = ½ of the total number of students.

Option D is TRUE.

Therefore, we can conclude that the statement, "A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie" is NOT TRUE.

3 0

3 years ago

Math pls help!! answers?

statuscvo [17]

Answer: Choice B) Infinitely many solutions

one solution: x = 8, y = -7/2, z = 0
another solution: x = -12, y = 13/2, z = 10

=======================================================

Explanation:

Here's the starting original augmented matrix.

$\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\-4 & 0 & -8 & -32\\\end{array}\right]$

We'll multiply everything in row 3 (abbreviated R3) by the value -1/4 or -0.25, which will make that -4 in the first column turn into a 1.

We use this notation to indicate what's going on: $(-1/4)*R3 \to R3$

That notation says "multiply everything in R3 by -1/4, then replace the old R3 with the new corresponding values".

So we have this next step:

$\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\1 & 0 & 2 & 8\\\end{array}\right]\begin{array}{l} \ \\\ \\(-1/4)*R3 \to R3\\\end{array}$

Notice that the new R3 is perfectly identical to R1.

So we can subtract rows R1 and R3, and replace R3 with the result of nothing but 0's

$\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l} \ \\\ \\R3-R1 \to R3\\\end{array}$

Whenever you get an entire row of 0's, it always means there are infinitely many solutions.

-------------------

Now let's handle the second row. That 5 needs to turn into a 0. We can multiply R1 by 5, and subtract that from R2.

So we need to compute 5*R1-R2 and have that replace R2.

$\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\0 & 1 & -1 & -7/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l} \ \\5*R1-R2 \to R2\ \\\ \\\end{array}$

Notice that in the third column of R2, we have 9-5*2 = 9-10 = -1. So we have -1 replace the 9. In the fourth column of R2, we have 73/2 - 5*8 = -7/2. So the -7/2 replaces the 73/2.

--------------------

At this point, the augmented matrix is in RREF form. RREF stands for Reduced Row Echelon Form. It seems a bit odd that the "F" of "RREF" stands for "form" even though we say "form" right after "RREF", but I digress.

Because the matrix is in RREF form, this means R1 and R2 lead to these equations:

$R1 : 1x+0y+2z = 8\\ R2: 0z+1y-1z = -7/2$

which simplify to

$R1: x+2z = 8\\R2: y-z = -7/2$

Let's get the z terms to each side like so:

$x+2z = 8\\x = -2z+8\\\text{ and }\\y-z = -7/2\\y = z-7/2\\$

Therefore, all of the solutions are of the form $(x,y,z) = (-2z+8, z-7/2, z)$ where z is any real number.

If z is allowed to be any real number, then we can simply pick any number we want to replace it. We consider z to be the "free variable", in that it's free to be whatever it wants. The values of x and y will depend on what we pick for z.

So the concept of "infinitely many solutions" doesn't exactly mean we can pick just any triple for x,y,z (admittedly it would be nice to randomly pick any 3 numbers off the top of my head and be done right away). Instead, we can pick anything we want for z, and whatever we picked, will directly determine x and y. The x and y are locked into place so to speak.

Let's say we picked z = 0.

That would lead to...

$x = -2z+8\\x = -2(0)+8\\x = 8\\\text{ and }\\y = z-7/2\\y = 0-7/2\\y = -7/2\\$

So z = 0 would lead to x = 8 and y = -7/2

Rearranging the items in alphabetical order gets us:

x = 8, y = -7/2, z = 0

We have one solution of (x,y,z) = (8, -7/2, 0)

Now let's say we picked z = 10

$x = -2z+8\\x = -2(10)+8\\x = -12\\\text{ and }\\y = z-7/2\\y = 10-7/2\\y = 13/2\\$

So we have x = -12, y = -13/2, z = 10

Another solution is (x,y,z) = (-12, 13/2, 10)

There's nothing special about z = 0 or z = 10. You can pick any two real numbers you want for z. Just be sure to recalculate the x and y values of course.

To verify each solution, you'll need to plug them back into the original equations formed by the original augmented matrix. After simplifying, you should get the same thing on both sides.

8 0

3 years ago

Hurry in test Brainliest!!!

tamaranim1 [39]

Answer:

y = (x - 4) - 4

Step-by-step explanation:

m = $\frac{y_{2} - y_{1} }{x_{2} -x_{1} }$