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sukhopar [10]
3 years ago
11

Need the answer to this question

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
3 0

Answer:

3 Rd one I think, it's correct

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Use the Distributive Property to solve the equation.
Arada [10]

Answer:

x= - 9

Step-by-step explanation:

  1. - 8(x +7) = 16
  2. -8x - 56 = 16
  3. -8x = 16 + 56
  4. -8x = 72
  5. x = 72 ÷ (-8)
  6. x = -9
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3 years ago
Maddie buys 3 sodas for $2 each. she pays with cash and receives $4 in change. write an equation that can be used to find d, the
Ierofanga [76]
3×2=6+4=d the 3×2 is how much the cost is and 6+4 is 10 so she paid 10 dollars
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3 years ago
How many solutions does this system have?
Luden [163]

Answer:

One solution

Step-by-step explanation:

i grapghed it the lines do intersect at one point

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3 years ago
Read 2 more answers
Rob Polanski is a tractor salesman. Last week his total sales amounted to $38,642.00, and he received $2,704.94 in commission. W
Alexeev081 [22]

To find the rate of commission, divide the amount of commission by his total sales:

2704.94 / 38642.00 = 0.07 = 7%

The answer is D. 7%

7 0
3 years ago
Read 2 more answers
The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machi
Sonja [21]

Answer:

t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

p_v =P(t_{(72)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=88.8 represent the sample mean

s=26.6 represent the sample standard deviation for the sample  

n=73 sample size  

\mu_o =93 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean i lower than 93 min, the system of hypothesis would be:  

Null hypothesis:\mu \geq 93  

Alternative hypothesis:\mu < 93  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=73-1=72  

Since is a one side test the p value would be:  

p_v =P(t_{(72)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

5 0
3 years ago
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