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suter [353]
4 years ago
11

HELP ME Solve. h ÷ 4 = 9

Mathematics
2 answers:
luda_lava [24]4 years ago
8 0

The answer would be

H÷4=9

   *4  *4

H=36

Umnica [9.8K]4 years ago
7 0
Isolate the h. Do the opposite of the sign given - Because of equal sign, what you do to one side you do to the other

h/4 =  9
h/4(4) = 9(4)
h = 9(4)
h = 36

36 is your answer for "h"

hope this helps


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4 &lt; 4x- 16<br> Please answer!
anzhelika [568]

Answer:

First you need to get your variable(x) by itself.

You do that by adding 16 to both sides of the greater than sign.

The 16 cancels or goes away. and 16 + 4 =20.

Your new equation is 20<4x

Finally you divide 20 by 4 to get 5

Your answer is 5<x or x>5.

6 0
3 years ago
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
4 years ago
A student used his place value chart to show a number. After the teacher instructed him to divide his number by 100,the chart sh
NeX [460]

When you divide by 100 you are essentially moving the decimal of the number two places to the left. In undoing this you would have to move the decimal of the number two places to the right.

28.003 would then turn into 2,800.3

Unfortunately I cannot draw a chart on here but that is the best I can do.

5 0
3 years ago
Read 2 more answers
Two major automobile manufacturers have produced compact cars with engines of the same size. We are interested in determining wh
Molodets [167]

Answer:

(A) The mean for the differences is 2.0.

(B) The test statistic is 1.617.

(C) At 90% confidence the null hypothesis should not be rejected.

Step-by-step explanation:

We are given that a random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance.

The following data (in miles per gallon) show the results of the test;

Driver         Manufacturer A               Manufacturer B

   1                      32                                       28

  2                      27                                       22

  3                      26                                       27

  4                      26                                       24

  5                      25                                       24

  6                      29                                       25

  7                       31                                       28

  8                      25                                       27

Let \mu_1 = mean MPG for the fuel efficiency of Manufacturer A brand

\mu_2 = mean MPG for the fuel efficiency of Manufacturer B brand

SO, Null Hypothesis, H_0 : \mu_1-\mu_2=0  or  \mu_1= \mu_2    {means that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles}

Alternate Hypothesis, H_A : \mu_1-\mu_2\neq 0  or  \mu_1\neq  \mu_2   {means that there is a significant difference in the mean MPG (miles per gallon) for the fuel efficiency of these two brands of automobiles}

The test statistics that will be used here is <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

                      T.S.  = \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t__n__1+_n__2-2

where, \bar X_1 = sample mean MPG for manufacturer A = \frac{\sum X_A}{n_A} = 27.625

\bar X_2 = sample mean MPG for manufacturer B =\frac{\sum X_B}{n_B} = 25.625

s_1 = sample standard deviation for manufacturer A = \sqrt{\frac{\sum (X_A-\bar X_A)^{2} }{n_A-1} } = 2.72

s_2 = sample standard deviation manufacturer B = \sqrt{\frac{\sum (X_B-\bar X_B)^{2} }{n_B-1} } = 2.20

n_1 = sample of cars selected from manufacturer A = 8

n_2 = sample of cars selected from manufacturer B = 8

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }   =  \sqrt{\frac{(8-1)\times 2.72^{2}+(8-1)\times 2.20^{2}  }{8+8-2} }  = 2.474

(A) The mean for the differences is = 27.625 - 25.625 = 2

(B) <u><em>The test statistics</em></u>  =  \frac{(27.625-25.625)-(0)}{2.474 \times \sqrt{\frac{1}{8}+\frac{1}{8}  } }  ~  t_1_4

                                     =  1.617

(C) Now at 10% significance level, the t table gives critical values between -1.761 and 1.761 at 14 degree of freedom for two-tailed test. Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles.

5 0
4 years ago
Which statements about square roots are true? Check all that apply.
Marina86 [1]
Bottom two are true
4 0
3 years ago
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