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Greeley [361]
3 years ago
14

What is the percent decrease from 95 to 68 round to the nearest percent

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
8 0

\qquad☀️ At first we need to find out the difference between 95 and 68.

\qquad\sf \longrightarrow  68 -95 \\

\qquad\sf \longrightarrow  -27

\qquad☀️ Now, divide the number by "the starting number.

\qquad\sf \longrightarrow \dfrac{-27 }{97}

\qquad\sf \longrightarrow -0.2842

\qquad☀️ And finally, we have to multiply the result by 100 to get it in terms of percent.

\qquad\sf \longrightarrow   -0.2842   \times 100

\qquad\red{\bf \longrightarrow   -28.42℅}

  • Hence, The answer will be -28.42℅

\qquad_________________________

Setler79 [48]3 years ago
6 0

The decrease is  (95-68) = 27 .

As a fraction, the decrease is  27/95  of the original amount.

To change any fraction to a decimal, do the division:

                            (27)  divided by  (95) = 0.2842...

To change any decimal to a percentage, move the

decimal point two places that ==> way:

                            0.2842...  =  28.42... %Answer:

Step-by-step explanation:

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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

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=[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.

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Step-by-step explanation:

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