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borishaifa [10]
3 years ago
10

1.) A,B,C,D? 2.) A,B,C,D? Please help mee

Mathematics
2 answers:
katrin [286]3 years ago
6 0
The first one is A.
The second one is C
Ludmilka [50]3 years ago
4 0

Answer:

1). A

2). C

Step-by-step explanation:

please mark brainliest

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Frank took a spelling test and got an 85%. If Frank got 34 questions correct, how many questions were on the test? Please be sur
Blizzard [7]

Answer:

40 questions

Step-by-step explanation:

34 / 85% = 40

40 * 85% = 34

5 0
3 years ago
3 dice of different colors are rolled, and the number coming up on each die is recorded. how many different outcomes are possibl
Dmitrij [34]
Assuming the dice go up to a maximum value of 6. The maximum number of different outcomes is 6^3 (6 times 6 times 6) which would be 216.
8 0
4 years ago
What is the sum of the first five terms of s geometric series with a1=6 and r=1/3
omeli [17]

S_{5} = \frac{242}{27} = 8 \frac{26}{27}

The sum of n terms in a geometric series is

S_{n} = \frac{a(1-r^n)}{1-r}, hence

S_{5} = \frac{6(1-(1/3)^5)}{1-1/3}

                        = \frac{6(1-1/243)}{2/3}

                        = 9 ( \frac{242}{243}) = \frac{242}{27}


7 0
4 years ago
Suppose a consumer product researcher wanted to find out whether a highlighter lasted longer than the manufacturer's claim that
rosijanka [135]

Answer:

C) t = 2.635; p = 0.006; Reject the null hypothesis; there is strong evidence to suggest that the highlighters last longer than 14 hours.

Step-by-step explanation:

Data given and notation  

\bar X=14.5 represent the sample mean

s=1.2 represent the sample standard deviation

n=40 sample size  

\mu_o =14 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 mean is higher than 14, the system of hypothesis would be:  

Null hypothesis:\mu \leq 14  

Alternative hypothesis:\mu > 14  

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{14.5-14}{\frac{1.2}{\sqrt{40}}}=2.635    

P-value

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

df=n-1=40-1=39  

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

p_v =P(t_{(39)}>2.635)=0.006  

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

The best option would be:

C) t = 2.635; p = 0.006; Reject the null hypothesis; there is strong evidence to suggest that the highlighters last longer than 14 hours.

4 0
3 years ago
Given P(E or F) = 0.16, P(E) = 0.12, and P(F) = 0.17, what is P(E and F)?
DiKsa [7]

P(E \cap F)=P(E)+P(F)-P(E \cup F)\\\\ P(E \cap F)=0.16+0.12-0.17=0.11

6 0
4 years ago
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