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2 years ago

Which algebraic expression could NOT represent the phrase below?

Mathematics

2 answers:

katrin2010 [14]2 years ago

7 0

Is B i hope it’s right

skad [1K]2 years ago

6 0

Answer:

Step-by-step explanation:

because 4 and 3 are grouped together you would add them first. this would change the outcome of the problem.

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To rent a jet ski at Sam’s costs $25 plus $3 per hour. At Claire’s, it costs $5 plus $8 per hour. At how many hours will the ren

BlackZzzverrR [31]

Answer: they will both cost $37 after 4 HOURS

step by step explanation:
3x - 25 = 8x + 5
- 3x. -3x

25 = 5x + 5
-5. -5

20 = 5x
/ 5. /5

X=4

Hope this helps

8 0

2 years ago

Evaluate the function f(x) = sec(x - ) for x = 0.

telo118 [61]

Answer:

explanation:

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3 years ago

A distribution center makes deliveries to food stores on a weekly basis. On the average, 50 stores are served daily. A typical s

patriot [66]

Answer:

3 doors needed

Step-by-step explanation:

The computation of the no of trucks doors that are needed is shown below:

= serve daily ÷ no of store orders ÷ no of store orders

= 50 ÷ 4 ÷ 4

= 3 doors needed

First we divided the serve daily from the no of store orders and then the value that comes so it would be divided by no of store orders again so that the no of truck doors could come

6 0

1 year ago

What is fig(-1)) for f(x) = 2x - 3 and g(x) = x² +11?

iVinArrow [24]

Answer:

retype the question or send a picture please

5 0

2 years ago

Suppose that the scores of bowlers in particular league follow a normal distribution such that the standard deviation of the pop

pav-90 [236]

Answer:

A 95% confidence interval for the population mean score for all bowlers in this league is [86.64, 94.48].

Step-by-step explanation:

Since in the question only 9 random scores are given, so I am performing the calculation using 9 random scores.

We are given that the scores of bowlers in particular league follow a normal distribution such that the standard deviation of the population is 6.

The accompanying data set of 9 random scores in ascending order is given as; 75, 86, 86, 88, 89, 93, 93, 98, 107

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean score = $\frac{\sum X}{n}$ = $\frac{815}{9}$ = 90.56

$\sigma$ = population standard deviation = 6

n = sample of random scores = 9

$\mu$ = population mean score for all bowlers

Here for constructing a 95% confidence interval we have used a One-sample z-test statistics because we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $90.56-1.96 \times {\frac{6}{\sqrt{9} } }$ , $90.56+1.96 \times {\frac{6}{\sqrt{9} } }$ ]

= [86.64 , 94.48]

Therefore, a 95% confidence interval for the population mean score for all bowlers in this league is [86.64, 94.48].

4 0

2 years ago