Hi can I have help? Please...

Spencer, Jaclyn, and Ayden go shrimping in

larisa [96]

The total pounds of shrimp they catch is 9.625 pounds

<h3>Sum of values and decimals</h3>

Decimal number are numbers that has decimal points. Given the following information;

The total shrimp Spencer catch = 2 pounds

The total shrimp Jaclyn catch = 3.25 pounds

The total shrimp Ayden catch = 4.375 pounds

Determine the total pounds of shrimp they catch

Total shrimp = 2 + 3.25 + 4.375

Total shrimp = 9.625 pounds

Hence the total pounds of shrimp they catch is 9.625 pounds

Learn more on sum of decimals here; brainly.com/question/1827193

#SPJ1

7 0

1 year ago

Y MORE THAN 4 is -36 translate for me please into math symbols then solve problem (list steps)

Sphinxa [80]

$y + 4 = -36

Subtract 4 both sides.

y = -40

Check solution. 

-40 + 4 = -36

This is correct.

You're welcome. 
$ Observe below please.

4 0

3 years ago

The set S consists of all odd multiples of 3: S = {..., -9, -3, 3, 9,...}. If the integers x and y are in S, then which of the f

tankabanditka [31]

I think it is gonna be xy

because if ur set contains all odd multiples of 3 (-9,-3,3,9,15,21,27...
and u pick any 2 of those numbers and multiply them u will end up with an odd numbered multiple of 3

and if u pick out any 2 numbers in ur set..and do any of the other things in ur answer choices, it will not always be an odd numbered multiple of 3

4 0

2 years ago

To what does the two-source rule refer?

Alla [95]

Step-by-step explanation:

p(A)={Ø,{2}{3},{2,3}

7 0

3 years ago

Sulfur compounds cause "off-odors" in wine, so winemakers want to know the odor threshold, the lowest concentration of a compoun

aev [14]

Answer:

a) $29.4-1.96\frac{7}{\sqrt{10}}=25.06$

$29.4+1.96\frac{7}{\sqrt{10}}=33.74$

So on this case the 95% confidence interval would be given by (25.06;33.74)

b) $z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988$

$p_v =P(z>1.988)=0.0234$

If we compare the p value and the significance level given $\alpha=Step-by-step explanation:Previous conceptsA confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". 
The margin of error is the range of values below and above the sample statistic in a confidence interval. 
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". 
Part a
Data given: 30 30 42 35 22 33 31 29 19 23
We can calculate the sample mean with the following formula:[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}= 29.4$ the sample mean

$\mu$ population mean (variable of interest)

$\sigma=7$ represent the population standard deviation

n=10 represent the sample size

95% confidence interval

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

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$29.4-1.96\frac{7}{\sqrt{10}}=25.06$

$29.4+1.96\frac{7}{\sqrt{10}}=33.74$

So on this case the 95% confidence interval would be given by (25.06;33.74)

Part b

What are H0 and Ha for this study?

Null hypothesis: $\mu \leq 25$

Alternative hypothesis : $\mu>25$

Compute the test statistic

The statistic for this case is given by:

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

z-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".

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

$z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988$

Give the appropriate conclusion for the test

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

$p_v =P(z>1.988)=0.0234$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

8 0

3 years ago