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Marta_Voda [28]
3 years ago
8

Please help with this question

Mathematics
1 answer:
AveGali [126]3 years ago
8 0

Answer:the image is not noticeable

Step-by-step explanation:

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Karen has $1.70 in coins. Karen has 8 coins, all of which are quarters or dimes.
kumpel [21]

x+y=8 and 25x+10y=170 are the linear equations.

x+y≤8 and 25x+10y≤170 are the inequalities.

Step-by-step explanation:

Given,

Worth of coins = $1.70 = 1.70*100 = 170 cents

Number of coins = 8

1 quarter = 25 cents

1 dime = 10 cents

Let,

x represent the number of quarters

y represent the number of dimes

1. Write an equation to represent the amount of coins Karen has.

x+y = 8

2.Write an equation to represent the value of the coins Karen has.

25x+10y=170

x+y=8 and 25x+10y=170 are the linear equations.

For inequalities, the amount cannot increase number of coins and worth but it can be less, therefore,

x+y≤8

25x+10y≤170

x+y≤8 and 25x+10y≤170 are the inequalities.

Keywords: linear equations, addition

Learn more about linear equations at:

  • brainly.com/question/10570041
  • brainly.com/question/10610299

#LearnwithBrainly

4 0
3 years ago
rename each mixed number as an equivalent mixed number with the same denominator. 4 6/9= 7 5/3= 3 7/8= 5 19/14=
denis-greek [22]
4 6/9= 38/9   
7 5/3= 26/3
3 7/8= 31/8
5 19/14 =89/14

5 0
3 years ago
Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg
Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let <em>X</em> = number of college women on a diet.

The probability of a woman being on diet is, P (X) = <em>p</em> = 0.70.

The sample of women selected is, <em>n</em> = 267.

The random variable thus follows a Binomial distribution with parameters <em>n</em> = 267 and <em>p</em> = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions (\hat p) follows a Normal distribution.  

The mean of this distribution is:

\mu_{\hat p} = p = 0.70

The standard deviation of this distribution is: \sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z

**Use the <em>z</em>-table for the probability.

P(\hat p \geq 0.75)=1-P(Z

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

7 0
3 years ago
A number less than 85 the number has 26 and 6 as factors
Len [333]
A number that is less than 85, the number has 26 and 6 as factors is,  78
7 0
3 years ago
A bag contains five green marbles three blue marbles and two red marbles if two consecutive marbles are drawn without replacemen
iren2701 [21]

Answer:

1 / 9

Step-by-step explanation:

P(Green) = 5 / 10

P(Blue) = 3 / 10

P(Red) = 2 / 10

These are independent events as one cannot occur at the same time as the other so can multiply their probabilities together

First choice green: 5 / 10

The denominator will decrease because we arent replacing the marbles after we take one out of the bag

Second choice red: 2 / 9

So

5 / 10 x 2 / 9

10 / 90 = 1 / 9

3 0
3 years ago
Read 2 more answers
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