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3 years ago

Divide 88 ballots into 2 groups so the ration is 1 to 3

2 answers:

8 0

The total parts desired is 1 + 3 = 4, therefore divide 88 by four to get one group, then multiply by 3 to get the other.
88 / 4 = 22
22 * 3 = 66
Final answer:
The two groups would have 22 and 66 ballots respectively.
Hope I helped :)

Lilit [14]3 years ago

4 0

No it would be 88 divided by 2 equal 44 out of 2 if that help brianliest answer plz and idk if u want that simplified or not

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Chase makes 2 gallons of soup for a dinner party. He serves 10 cups of soups to his guests. How many cups of soup will he have l

iVinArrow [24]

1 gallon = 16 cups

2 gallons = 16 x 2 = 32

32-10=22

there will be 22 cups left over

4 0

3 years ago

Read 2 more answers

Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longbo

Alisiya [41]

Answer:

Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0: μL > μs against the claim Ha: μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/ √s1/n1+ s2/n2

1) Calculations

Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58) ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

6 0

3 years ago

Rita has already prepared 3 kilograms of dough and will continue preparing 1 kilogram of dough every hour. How much dough did Ri

Digiron [165]

Answer:

Step-by-step explanation:

9+3=12

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3 years ago

If you borrowed $3,500 to buy a new car at 3% interest and paid it back in 4 years. How much would you pay in interest?

tankabanditka [31]

Answer:

$420

Step-by-step explanation:

Calculate 3% of 3,500:

0.03*3500 = 105

105 per year for 4 years:

105*4 = 420

7 0

3 years ago

The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day

Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(between 236 and 281 days)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%$

b) a) P(last between 236 and 296)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%$

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least $1-\dfrac{1}{k^2}$ data lies within k standard deviation of mean.

For k = 2

$1-\dfrac{1}{(2)^2} = 75\%$

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0

3 years ago