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4 years ago

7

Using the following to answer the question: p: 3 is an even number q: 9 is an even number Using a truth table, determine whether

the following conjunction is true or false: p ^ q

1 answer:

kogti [31]4 years ago

8 0

false.............................

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nignag [31]

B and c would be answered

7 0

4 years ago

Exponent of 9540000 using power of 10

xenn [34]

Are you doing scientific notation or expanded form?
Scientific notation:
9.54×10^6
Expanded Form:
9×10^6+5×10^5+4×10^4

5 0

3 years ago

A) a shopkeeper buys a camera for $300 and sells it at $360. Calculate the percentage profit.

Lilit [14]

A) profit/original price x100 =percentage profit

(Profit: 360-300=$60)

=60/300 x100
=20%

b) two cameras (original price): 300x2= $600
two cameras (price sold): 360x2 = $720

Profit without discount: 720-600= $120

120-100= $20 discount

20/720 x100 =2.78%

4 0

2 years ago

Read 2 more answers

Please help! Look at the picture.

monitta

Answer:

Step-by-step explanation:

(27)^(-1/3) + (32)^(-2/5)

1/3 + 1/4

4/12 + 3/12 = 7/12

8 0

3 years ago

Suppose that textbook weights are normally distributed. You measure 33 textbooks' weights, and find they have a mean weight of 7

AleksAgata [21]

Answer:

$75-2.58\frac{13.3}{\sqrt{33}}=69.027$

$75+2.58\frac{13.3}{\sqrt{33}}=80.793$

And the 95% confidence interval would be between (69.027;80.793)

Step-by-step explanation:

Information given

$\bar X=75$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=13.3$ represent the population standard deviation

n=33 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are given by:

$df=n-1=33-1=32$

The Confidence level is 0.99 or 99%, the significance would be $\alpha=0.01$ and $\alpha/2 =0.005$ , the critical value for this case would be $z_{\alpha/2}=2.58$

And replacing we got:

$75-2.58\frac{13.3}{\sqrt{33}}=69.027$

$75+2.58\frac{13.3}{\sqrt{33}}=80.793$

And the 95% confidence interval would be between (69.027;80.793)

6 0

3 years ago