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3 years ago

11

Juan is saving money for a down payment on a car. He has already saved $300 and plans to continue saving $35 each week. How many

weeks will it take Juan to save $1,000? (I also need the equation you used to solve it, thank you).

1 answer:

Lemur [1.5K]3 years ago

8 0

1000=300+35x

so 20 weeks

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Karina’s paycheck stub had several items listed on it. Which of these items listed on her paycheck is not a deduction?

Katyanochek1 [597]

Answer:I believe it’s tuition reimbursement edge 21

Step-by-step explanation:

3 0

3 years ago

Malcolm Little High School has sopho- mores, juniors, and seniors, but no freshmen. The ratio of sophomores to juniors is 5 to 4

djverab [1.8K]

Answer:

161 seniors

Step-by-step explanation:

According to the problem, given data are as follows:

Total number of sophomores = 300

Sophomores to junior ratio = 5:4

So, Number of Juniors = 300 × (4 ÷ 5)

= 240

Now, Juniors to senior ratio = 3:2

So, we can calculate the number of seniors by using following method:

Number of seniors = 240 × ( 2 ÷ 3)

= 240 × 0.67

= 160.8 or 161

6 0

3 years ago

If the ríbbon costs 10 cents per foot what is the total cost of ribbon in dollars

True [87]

The answer will be 10 ft because its 10 cents per foot

4 0

4 years ago

A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen

tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average desired retirement age = 55 years

$\sigma$ = sample standard deviation = 3.4 years

n = sample of seniors = 101

$\mu$ = true mean retirement age of all college students

<em>Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 96% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.114 < $t_1_0_0$ < 2.114) = 0.96 {As the critical value of t at 100 degree

of freedom are -2.114 & 2.114 with P = 2%}

P(-2.114 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.114) = 0.96

P( $-2.114 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

<u>96% confidence interval for</u> $\mu$ = [ $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $55-2.114 \times {\frac{3.4}{\sqrt{101} } }$ , $55+2.114 \times {\frac{3.4}{\sqrt{101} } }$ ]

= [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

7 0

3 years ago

Suppose 15 quarts of a solution that is 35% antifreeze is mixed with 5 quarts of a solution that is 72% antifreeze. 1.how many q

Finger [1]

Answer:

1. 8.85 quarts

2. 44.25%

Step-by-step explanation:

In 15 quarts of a solution with percentage of antifreeze = 35%

Amount of antifreeze = 35% × 15

= 0.35 × 15

= 5.25 quarts

that solution is mixed with 5 quarts of a solution with percentage of antifreeze = 72%

Amount of antifreeze = 72% × 5

= 0.72 × 5

= 3.6 quarts

Total amount of mixture = 15 quarts + 5 quarts = 20 quarts

1. Now we will calculate the total amount of antifreeze in the resulting mixture.

= 5.25 + 3.6 = 8.85 quarts

2. The percentage of the resulting mixture is antifreeze

= $\frac{8.85}{20}\times 100$

= 44.25%

1. total amount of antifreeze is 8.85 quarts

2. the percentage of antifreeze is 44.25%

3 0

3 years ago