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Alexxandr [17]
2 years ago
11

Isabella's teacher said she must read more than 70 pages of her book by the end of the week. Isabella is able to read 12 pages o

f her book per hour, and she has already read 10 pages.
Which statement describes how many more hours Isabella must read her book to complete the assignment?

She must read up to 6 more hours.

She must read more than 5 more hours.

She must read fewer than 5 more hours.

She must read no less than 6 more hours.
Mathematics
1 answer:
VARVARA [1.3K]2 years ago
3 0

Answer:

6

Step-by-step explanation:

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70600

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I will give out BRAINLIEST for the correct answer!
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Answer:

2

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3 years ago
Help me please I really need help
jeka94

Answer:

A. The y-intercept of g(x) is less than the y-intercept of f(x).

Step-by-step explanation:

The x-intercept, or when x = 0, of f(x) is -4, the x-intercept of g(x) is -8, so g(x) are neither greater nor equal to f(x), this marks out C and D. The y-intercept, or when y = 0, is in this case f(x) or g(x). The y-intercept of f(x) is 16, the y-intercept of g(x) is 4, so the y-intercept of g(x) is not equal to the y-intercept of f(x), this marks out B. Now to check A, 4 < 16, so y-intercept of g(x) < y-intercept of f(x), the answer is A

5 0
2 years ago
Im solving variables i need to simplify -5x + 3 (x+5) = -3
victus00 [196]

Answer:

x=9

Step-by-step explanation:

Add similar elements: -5x+3x=-2x

-2x+15=-3

Subtract 15 from both sides

-2x + 15-15=-3-15

Simplify

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The amount of money spent on textbooks per year for students is approximately normal.
Ostrovityanka [42]

Answer:a

a

   336.04    <  \mu < 443.96

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  0.4107 <  p  < 0.4293

Step-by-step explanation:

From the question we are  told that

   The sample size  n =  19

    The sample mean is  \= x  = \$\  390

    The  standard deviation is  \sigma =  \$ \  120

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha = 100 -  95

          \alpha  =  5 \%

          \alpha  =  0.05

Next we obtain the critical value of \frac{\alpha }{2} from the normal distribution table

    So  

         Z_{\frac{\alpha }{2} } =  1.96

The  margin of error is mathematically represented as

      E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>    E = 1.96 *  \frac{120}{\sqrt{19} }

=>   E = 53.96

The 95% confidence interval is  

     \= x  -  E  <  \mu < \= x  +  E

=>   390  -   53.96   <  \mu < 390  -   53.96

=>  336.04    <  \mu < 443.96

When the confidence level increases the Z_{\frac{\alpha }{2} } also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         n  \  \ \alpha  \ \  \frac{1}{E^2 }

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

       \r p  =  \frac{210}{500}

       \r p  = 0.42

Given that the confidence level is 0.99 the level of significance is  \alpha =  0.01

The critical value of \frac{\alpha }{2} from the normal distribution table is  

      Z_{\frac{\alpha }{2} }  =  2.58

  Generally the margin of error is mathematically represented as

       E =  Z_{\frac{\alpha }{2} }*  \sqrt{ \frac{\r p (1- \r p )}{n} }

=>   E =  0.42 *  \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }

=>     E =  0.0093

The 99% confidence interval  is

     \r p  -  E <  p  < \r p  +  E

     0.42  -  0.0093 <  p  < 0.42  +  0.0093

     0.4107 <  p  < 0.4293

 

4 0
3 years ago
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