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gogolik [260]
2 years ago
14

Erica solved the equation −5x − 25 = 78; her work is shown below. Identify the error and where it was made.

Mathematics
1 answer:
Svetlanka [38]2 years ago
6 0
Step 1 A is correct.....
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Please help!!!!!!!!!!!!!!!
icang [17]

Answer:

im not sure

Step-by-step explanation:

go to www.analyzemath.com

4 0
3 years ago
A rectangle has a perimeter of 32 in. find the length and width of the rectangle under which the area is the largest. follow the
Dafna1 [17]
Let width and length be x and y respectively.
Perimeter (32in) =2x+2y=> 16=x+y => y=16-x
Area, A = xy = x(16-x) = 16x-x^2
The function to maximize is area: A=16 x-x^2
For maximum area, the first derivative of A =0 => A'=16-2x =0
Solving for x: 16-2x=0 =>2x=16 => x=8 in
And therefore, y=16-8 = 8 in
5 0
3 years ago
Evaluate x-y/5 when x is 23 more than 3 times y and y=6
posledela

Answer:

7

Step-by-step explanation:

1) "x is 23 more than 3 times y" is the same thing as "x = 23 + 3y"

REMEMBER: when you see "is" replace is with an equals sign.

2) Solve for x with your new equation.

3) Plug 6 into the y ("y = 6")

x = 23 + 3 x 6... x = 23 + 18... x = 41

4) Plug in your variables, and evaluate.

(41 - 6)/5... 35/5... 7!

8 0
3 years ago
Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o
Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0
2 years ago
According to the Rational Root Theorem, what are all the potential rational roots of f(x) = 15x11 – 6x8 + x3 – 4x + 3?
scoundrel [369]

we have

f(x) = 15x^{11} -6x^{8} + x^{3} - 4x + 3

we know that

<u>The Rational Root Theorem</u> states that when a root 'x' is written as a fraction in lowest terms

x=\frac{p}{q}

p is an integer factor of the constant term, and q is an integer factor of the coefficient of the first monomial.

So

in this problem

the constant term is equal to 3

and the first monomial is equal to 15x^{11} -----> coefficient is 15

So

possible values of p are 1, and\ 3

possible values of q are 1, 3, 5, and\ 15

therefore

<u>the answer is</u>

The all potential rational roots of f(x) are

(+/-)\frac{1}{15},(+/-)\frac{1}{5},(+/-)\frac{1}{3},(+/-)\frac{3}{5},(+/-)1,(+/-)3


3 0
3 years ago
Read 2 more answers
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