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koban [17]
2 years ago
8

Anyone know the answer and the check to the equation 7 - 4c = 3c - 7 ?

Mathematics
2 answers:
givi [52]2 years ago
8 0

Answer:

c= 2

Step-by-step explanation:

7 - 4c = 3c - 7

Add 4c to both sides

7 = 7c - 7

Add 7 to both sides

14 = 7c

Divide both sides by 7

2 = c

Gelneren [198K]2 years ago
3 0

Answer:

c=2

Step-by-step explanation:

7-4c=3c-7

7+7=3c+4c

14=7c

Dividing both sides by 7

c=2

CHECK:

Putting c in the above equation

7-4(2)=3(2)-7

7-8=6-7

-1=-1

Hence the above solution satisfies the given equation

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At the end of a​ party, 3/4 cup of dip is left. Jim divides 4/7 of the leftover dip equally between 2 friends. How much dip does
Vesnalui [34]

Answer:

First, we need to find the amount of dip that was divided amount the two friends.

We know that it is 4/7 of the left dip  

therefore:

amount divided among the two friends = (4/7) x (3/4) = 3/7 of the original amount of dip.

This amount is divided among two friends,

therefore:

amount that each friend gets = (3/7) / (2) = 3/14 = 0.2 of the amount of dip

tell me if this is completely wrong because i'm not good at this

7 0
3 years ago
You go out to eat with your family. The subtotal on your bill is $45.90. You must pay a 8% sales tax AND you would like to leave
Eddi Din [679]

Answer:

The answer is 57.83

Step-by-step explanation:

find 8% of 45.90 and add 18% of 45.90 and add them together.

8%*45.90=3.67

18%*45.90=8.26

Add, 45.90+8.26+3.67

The final price of your meal is 57.83

7 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
I really need help! Please, please, PLEASE do not give me a link.
MArishka [77]

Answer:

7 fiction and 23 non- fiction

Step-by-step explanation:

The number of non-fiction read was 5 less than 4 times the number of fiction books.

Assume fiction books are x.

An equation representing this would be:

x + (4x - 5) = 30

5x - 5 = 30

5x = 30 + 5

x = 35 / 5

x = 7 fiction books

Non-fiction books:

= 4x - 5

= 4 * 7 - 5

= 23 non - fiction books

3 0
2 years ago
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2. Sequoia trees can grow to more than 90 m in
sashaice [31]

Answer: F

Step-by-step explanation: 90/3.28 equals 27.4390244

5 0
3 years ago
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