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3 years ago

Write a numerical expression for the phrase “16 times the difference of 9 and 3.” What operation should you perform first

Mathematics

1 answer:

amid [387]3 years ago

5 0

16(9-3)
Subtract what is in the parentheses then multiply that answer by 16, 6•16=96

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Please answer this correctly

Stolb23 [73]

Answer:

Jenny swam more laps.

Step-by-step explanation:

Lets take this step by step. If Edith swam fewer laps than Shira but more than Gemma, then G < E < S. If Jenny swam more than Shira, then that means Gemma swam the fewest, then Edith, then Shira, and finally Jenny who swam the most laps.

3 0

3 years ago

Read 2 more answers

Which shows 52² - 42² being value added using the difference of perfect squares method

8090 [49]

Answer:

52² - 42² = ( 52 - 42 )· ( 52 + 42 ) = 10 · 94 = 940;

Step-by-step explanation:

4 0

3 years ago

(20 Poyntz) This question is UNSOLVABLE find out for yourself on this IMPOSSIBLE QUESTION

Pavlova-9 [17]

Answer:

1st one

Step-by-step explanation:

-3 is ur starting point and u go up 1 to the side 2

8 0

3 years ago

Read 2 more answers

A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of

Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

Hours of TV per week (X): 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

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

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$ = 4.61

n = sample of people = 20

$\mu$ = true average number of hours of TV watched per week

Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the true average, $\mu$ is ;

P(-1.33 < $t_1_9$ < 1.33) = 0.80 {As the critical value of t at 19 degrees of

freedom are -1.33 & 1.33 with P = 10%}

P(-1.33 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

= [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

7 0

3 years ago

Nora read 1/4 of a book on monday and 2/5 on tuesday. what fraction of the book does she have left to read?

taurus [48]

On Monday, Nora read ¼ of the book, therefore she is left with ¾ of the original book.

Then on Tuesday, she read 2/5 of what was left on Monday therefore she is left with 3/5 of ¾ of the original book. This is equal to:

(3/5) (3/4) = 9/20

She still needs to read 9/20 of the book or 0.45 or 45% of the book.

7 0

3 years ago