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nadezda [96]
3 years ago
12

Please help me at this math proplem thank you pleas help me

Mathematics
1 answer:
Shkiper50 [21]3 years ago
4 0

which math problem?? There is many on the picture

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What is the result of 72 divided by 12? 6 8 9 12
galben [10]

Answer:

the answer is 6

Step-by-step explanation:

7 0
3 years ago
The product of 4 and a number increased by 7,is -36. translate to an equation then solve
evablogger [386]
You have an English language sentence that we need to translate into Mathematics language.
English)  (the product of a number and 9), (increased by 4), (is 58) Math)     (Number x 4), (+ 7), (= -36) I put in the () to show you what parts go togetherCall the Number N, we can use anything (x,s, ...) but N makes sense (N x 4) + 7 = -36 Can you do the rest?  subtract 7 from both sides, divide both sides my 4, there is the answer

5 0
4 years ago
How many questions were on Victoria test
Naddik [55]
25!

(80/100)*x=20
x=25

100% Verified

Hope this helps! :)
7 0
3 years ago
Super Bowl XLVI was played between the New York Giants and the New England Patriots in Indianapolis. Due to a decade-long rivalr
Diano4ka-milaya [45]

Answer:

Probability that from a sample of 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.

Step-by-step explanation:

We are given that Due to a decade-long rivalry between the Patriots and the city's own team, the Colts, most Indianapolis residents were rooting heartily for the Giants. Suppose that 90% of Indianapolis residents wanted the Giants to beat the Patriots.

Let p = % of Indianapolis residents wanted the Giants to beat the Patriots = 90%

The z-score probability distribution for proportion is given by;

                   Z = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where,  \hat p = % of Indianapolis residents who were rooting for the Giants in Super Bowl XLVI in a sample of 200 residents = \frac{170}{200} = 0.85

           n = sample of residents = 200

So, probability that from a sample of 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLVI is given by = P(\hat p < 0.85)

     P(\hat p < 0.85) = P( \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < \frac{0.85-0.90}{\sqrt{\frac{0.85(1-0.85)}{200} } } ) = P(Z < -1.98) = 1 - P(Z \leq 1.98)

                                                                   = 1 - 0.97615 = 0.02385

<em>The above probability is calculated using z table by looking at value of x = 01.98 in the z table which have an area of 0.97615.</em>

<em />

Therefore, probability that fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.

7 0
3 years ago
a President Bush international and 2005 the newspapers headlines started about four thousand million people in it in the newspap
svlad2 [7]
The smallest would be 400,000 and the largest would be the same because 0 can't round up or down
4 0
3 years ago
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