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Luden [163]
3 years ago
14

Estimate and then compute each product. Check that your answer is reasonable. 531 X 4 7

Mathematics
1 answer:
zepelin [54]3 years ago
7 0

Answer:

25,000

Step-by-step explanation:

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A store is having a sale on walnuts and chocolate chips. For 9 pounds of walnuts and 7 pounds of chocolate chips, the total cost
vfiekz [6]

Answer:

Walnuts = $1.75 per pound

Choc chips = $3.75 per pound

Step-by-step explanation:

In photo, is working out.

6 0
4 years ago
Find the missing variable for a triangle: a=48insq, h=?, b=12 in
Aleksandr [31]
A^2+B^2=C^2
48^2+12^2=C^2
2304+144=C^2
2448=C^2
now you keep this in radical form or you can round your answer.
Rounded=== 49.48=C
Radical==== 12√17=C
6 0
4 years ago
Can someone help me with number 2,4,6,8,10,12,14,16,18,20 please??
nignag [31]

Answer:

Step-by-step explanation:

  • 17= 8+9
  • 42=3*14
8 0
3 years ago
Will mark as brainliest if correct
stellarik [79]

Answer:

-2/3

Step-by-step explanation:

The only way the answer could be one is if the exponent is 0

So 3x-2=0:

3x = -2

x = -2/3

Hope this helps!

8 0
4 years ago
An automobile manufacturer has given its car a 52.6 miles/gallon (MPG) rating. An independent testing firm has been contracted t
Arte-miy333 [17]

Answer:

Null hypothesis:\mu = 52.6  

Alternative hypothesis:\mu \neq 52.6  

z=\frac{52.8-52.6}{\frac{1.6}{\sqrt{250}}}=1.976    

p_v =2*P(z>1.976)=0.0482  

Since the p value is lower than the significance level wedon't have enough evidence to conclude that the true mean is significantly different from 52.6 MPG.

Step-by-step explanation:

Information provided

\bar X=52.8 represent the sample mean  for the MPG of the cars

\sigma=1.6 represent the population standard deviation

n=250 sample size  of cars

\mu_o =52.6 represent the value that we want to test

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

z would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis

We need to conduct a hypothesis in order to check if the true mean of MPG is different from 52.6 MPG, the system of hypothesis would be:  

Null hypothesis:\mu = 52.6  

Alternative hypothesis:\mu \neq 52.6  

Since we know the population deviation the statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Calculate the statistic

Replacing we have this:

z=\frac{52.8-52.6}{\frac{1.6}{\sqrt{250}}}=1.976    

Decision

Since is a two tailed test the p value would be:  

p_v =2*P(z>1.976)=0.0482  

Since the p value is lower than the significance level wedon't have enough evidence to conclude that the true mean is significantly different from 52.6 MPG.

3 0
3 years ago
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