All categories

4 years ago

Two-thirds of a number plus 17 is at least 29

Mathematics

1 answer:

Mandarinka [93]4 years ago

7 0

Answer:

Step-by-step explanation:

If you do 2/3 of the number three you have 1 and 1 + 17 = 18

You might be interested in

Solve the following equation -2y-6=4

LenKa [72]

Answer:

Y= -5

Explanation: have an amazing day!!

3 0

3 years ago

Which of the following is a solution to the equation y=3x - 1

kirill [66]

When x is 4, 

y = 11 (11 =/= 4) 

When x is 2, 

y = 5 (This is correct) 

When x is 4, 

y = 11 (11 =/= 3) 

When x is 0, 

y = -1 (-1 =/= -3)

8 0

3 years ago

Step by step instructions

alisha [4.7K]

Since there is a multiplier of 7 on the left, and both terms on the right are divisible by 7, I'd start by dividing the equation by 7

|8 - 3x| = 3x - 7

The absolute value function may or may not negate its argument. (The argument is negated if it is negative.) Hence this can resolve to two equations

-(8 -3x) = 3x -7

(8 -3x) = 3x -7

_____

Starting with the first of these, we can eliminate parentheses to get

... -8 +3x = 3x -7

Subtracting 3x from both sides gives the FALSE statement

... -8 = -7 . . . . FALSE

Thus, this version of the equation has No Solution.

Looking at the second of the equations above, we can add 3x to get

... 8 = 6x -7

Adding 7 and dividing by 6 gives

... (8 +7)/6 = x = 5/2

The solution is x = 5/2.

3 0

4 years ago

How many solutions does the system of equations have? x = -3y + 4 6y + 2x = 8

Setler79 [48]

Answer:

The equation x = -3y + 4 6y + 2x = 8 has infinite number of solutions.

7 0

3 years ago

I need to know how to do the whole thing and understand it.

patriot [66]

We are given the data on the number of candies handed by neighborhood A and neighborhood B.

Let us first find the mean and variance of each neighborhood.

Mean:

$\bar{x}_A=\frac{\sum x}{N_1}=\frac{12}{6}=2$ $\bar{x}_B=\frac{\sum x}{N_2}=\frac{20}{6}=3.33$

Variance:

$s_A^2=\frac{\sum x^2}{N_1}-\bar{x}_A^2=\frac{28}{6}-2^2=0.667$ $s_B^2=\frac{\sum x^2}{N_2}-\bar{x}_B^2=\frac{80}{6}-3.33^2=2.244$

A. Null hypothesis:

The null hypothesis is that there is no difference in the mean number of candies handed out by neighborhoods A and B.

$H_0:\;\mu_A=\mu_B$

Research hypothesis:

The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.

$H_a:\;\mu_A>\mu_B$

Test statistic (t):

The test statistic of a two-sample t-test is given by

$t=\frac{\bar{x}_A-\bar{x}_B}{s_p}$

Where sp is the pooled standard deviation given by

$\begin{gathered} s_p=\sqrt{\frac{N_1s_1^2+N_2s_2^2}{N_1+N_2-2}(\frac{N_1+N_2}{N_1\cdot N_2}}) \\ s_p=\sqrt{\frac{6\cdot0.667+6\cdot2.244}{6+6-2}(\frac{6+6}{6\cdot6})} \\ s_p=0.763 \end{gathered}$ $t=\frac{2-3.33}{0.763}=-1.74$

So, the test statistic is -1.74

Critical t:

Degree of freedom = N1 + N2 - 2 = 6+6-2 = 10

Level of significance = 0.05

The right-tailed critical value for α = 0.05 and df = 10 is found to be 1.81

Critical t = 1.81

We will reject the null hypothesis because the calculated t-value is less than the critical value.

Interpretation:

This means that we do not have enough evidence to conclude that neighborhood A gives out more candies than neighborhood B.

6 0

1 year ago