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Neporo4naja [7]
3 years ago
12

Jennifer has 8 bags of marbles. Each bag has 5 marbles in it. If she divides all of the marbles into 4 groups, how many will be

in
each group?
Mathematics
2 answers:
Mrac [35]3 years ago
8 0

Answer:

Step-by-step explanation:

8 x 5=40

40 ÷ 4=10

hope this helps!

emmasim [6.3K]3 years ago
3 0

the anwers to the problem is 8,5,4

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Im stuck on this question helm me out I will mark you as brainliest​
agasfer [191]

Answer: <em><u>it is =4176000000000000</u></em>

Step-by-step explanation:

(2.9)(100000)(7.2)(10^2)

5(10^−8)

=

(290000)(7.2)(10^2)

5(10^−8)

=

2088000(10^2)

5(10^−8)

=

(2088000)(100)

5(10^−8)

=

208800000

5(10^−8)

=

208800000

5(1/100000000)=

208800000/1

20000000

=4176000000000000

hope i helped

-lvr

8 0
3 years ago
Factor 30b2 + 48b - 24
iogann1982 [59]
IM going to assume that the "2" is to be squared. If so, then the answer is 2(15x^2+8)
6 0
3 years ago
?????????????? So like anyone know this?
Ivanshal [37]
Not sure sorry I’m not good at math need help with something else..
8 0
3 years ago
PLEASE HELP ME I BEG OF YOU IF YOU HELP ME I WILL GIVE YOU BRAINTLIEST"Julie goes to the sports store and spends $40.50 before t
Pie

Answer:

6.25

Step-by-step explanation:

Plz give me brainleist like u said.

3 0
3 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha 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 true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

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

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
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