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Whitepunk [10]
3 years ago
14

6300/90???? PLS helpppp

Mathematics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer: 70

Step-by-step explanation: 6300/90=70 Brainliest please?

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3 x 4/7 =.Pls help me fast i need the answer
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Answer: 1.71428571429

Step-by-step explanation:

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Lydia is filling soda machines with soda. Each machine should get 25 cans . If she has 140 sodas and fills all but the last mach
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The last machine will only have 15 sodas
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Points A and B on the coordinate grid below show the positions of two midfield players of a soccer team:
ehidna [41]
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4 years ago
What is the sum of the arithmetic sequence 3, 9, 15..., if there are 36 terms?
EastWind [94]
We know that
the formula of the <span>the sum of the arithmetic sequence is
Sum=n*(a1+an)/2

n=36
a1=3
a36=?

find a36
an=a1+(n-1)*d

for n=2   a2=9
9=3+(2-1)*d----> 9=3+d----> d=6
so
for n=36
a36=a1+(36-1)*6----> a36=3+35*6----> a36=213

</span>Sum=n*(a1+an)/2-----> 36*[3+213]/2----> 3888

the answer is
3888
8 0
4 years ago
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If a hurricane was headed your way, would you evacuate? The headline of a press release states, "Thirty-four Percent of People o
IgorC [24]

Answer:

The 98% confidence interval would be given by (0.326;0.354)

Step-by-step explanation:

1) Notation and definitions

X=63 number of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

n=6138 random sample taken

\hat p=0.34 estimated proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

p true population proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

A 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".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

2) Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by \alpha=1-0.98=0.02 and \alpha/2 =0.01. And the critical value would be given by:

z_{\alpha/2}=-2.33, z_{1-\alpha/2}=2.33

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

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.34 - 2.33\sqrt{\frac{0.34(1-0.34)}{6138}}=0.326

0.34 + 2.33\sqrt{\frac{0.34(1-0.34)}{6138}}=0.354

The 98% confidence interval would be given by (0.326;0.354)

8 0
3 years ago
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