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salantis [7]
2 years ago
15

The numbers 1 through 25 are written on 25 cards with onenumber on each card. Sara picks one of the 25 cards at random.What is t

he probability that the number on her card will be amultiple of 2 or 5
Mathematics
1 answer:
nydimaria [60]2 years ago
5 0

Answer:

17/25

Step-by-step explanation:

sorry if I'm wrong

lay them out on.a piece of paper and circle every multiple of 2

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Which of the equations shows an application of the Zero Property of Multiplication? A. 25+0=25 B. 13(3)= C. 0∙2=0 D. (34−2)−(34+
Montano1993 [528]

Answer:

0∙2=0

Step-by-step explanation:

Zero Property of multiplication is as follows :

The product of a number and 0 is equal to 0.

In option (c), 0 is multiplied by 2 and as a result, we get 0 as the answer.

It means,

0∙2=0

Hence, option (c) i.e. 0∙2=0 shows the Zero Property of Multiplication.

5 0
3 years ago
You have $18 to buy peppers. Peppers cost $1.50 each. Write and solve and inequality that represents the number of peppers you c
IrinaVladis [17]

Answer:

1.50x=18

Step-by-step explanation:

you need to figure out how many you can buy, so we will put x in there and it will all equal 18, therefore heres that equal sign. so yeh.

3 0
3 years ago
Read 2 more answers
Find the critical value of x2 based on the given information H1 >3.5 n=14 a=0.05
egoroff_w [7]

Answer:

The degrees of freedom are given by:

df = n-1=14-1=13

The significance level is \alpha=0.05 and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

\chi^2_{\alpha}= 22.362

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

Step-by-step explanation:

We have the followign system of hypothesis:

Null hypothesis: \sigma^2 \leq 3.5

Alternative hypothesis: \sigma^2 >3.5

The degrees of freedom are given by:

df = n-1=14-1=13

The significance level is \alpha=0.05 and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

\chi^2_{\alpha}= 22.362

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

4 0
3 years ago
Which of these expressions CANNOT be simplified by combining like terms?
densk [106]
5ab^{3} +7-3a^{2} b ^{2} +a^{3}b+10 can be simplified to by adding the 7 and 10 to get a^{3}b-3a^{2} b ^{2}+ 5ab^{3} + 17.

5ab^{3}+3a^{2}b^{2}+a^{3}b -10 cannot be simplified any more by combining like terms.

By distributing the 2b into the parentheses, you can simplify the expression:
5ab^{3}+2b(3ab^{2})+a^{3}b-10\\
5ab^{3}+6ab^{3}+a^{3}b-10\\
11ab^{3}+a^{3}b-10

Here you can just add:
5a^{3}b^{3}+3a^{2}b^{2}+a^{3}b^{3}-10ab\\
6a^{3}b^{3}+3a^{2}b^{2}-10ab

Thus, the only expression that cannot simplify any more using adding like terms is the second, 5ab^{3}+3a^{2}b^{2}+a^{3}b -10.
3 0
3 years ago
Read 2 more answers
5/8 - 3/4(8 - 1/3) + 1
Anon25 [30]
The answer is definitely -4 1/8
6 0
2 years ago
Read 2 more answers
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