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3 years ago

John does 96 pushups and 4 days how many pushups does he do in one day

Mathematics

2 answers:

Vsevolod [243]3 years ago

4 0

96 divided amongst 4 days, 96 divided by 4 is 24.

SVETLANKA909090 [29]3 years ago

3 0

This answer can be found by taking 96/4 (96 decided by 4), which is 24. John does 24 push ups per 1 day.

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PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

Mila [183]

Answer:

$m\angle H = 44.4\°$ .

Step-by-step explanation:

Given:

In Right Angle Triangle GIH

∠ I = 90°

GI = 7 ....Side opposite to angle H

GH = 10 .... Hypotenuse

To Find:

m∠H = ?

Solution:

In Right Angle Triangle ABC ,Sine Identity,

$sin \ H = \frac{Oppsite\ side\ to\ \angle H}{Hypotenuse}$

Substituting the values we get;

$sin\ H = \frac{7}{10} = 0.7$

Now taking $sin^{-1}$ we get;

$\angle H = sin^{-1}\ 0.7 = 44.427$

rounding to nearest tenth we get.

$m\angle H = 44.4\°$ .

Hence $m\angle H = 44.4\°$ .

6 0

2 years ago

3.11 A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of 3 of the sets. If x is the nu

djverab [1.8K]

Answer:

Probability distribution for x:

$P(x=0)=0.3644\\\\P(x=1)=0.4373\\\\P(x=2)=0.1749\\\\P(x=3)=0.0233\\\\$

Step-by-step explanation:

We can model the number of defective sets in the group of TV sets (variable x) as a binomial variable, with sample size=3 and probability of success p=2/7≈0.2857.

The probability of k defective sets in the group is:

$P(x=k) = \dbinom{n}{k} p^{k}q^{n-k}$

So, we have this probabilty distribution for x:

$P(x=0) = \dbinom{3}{0} p^{0}q^{3}=1*1*0.3644=0.3644\\\\\\P(x=1) = \dbinom{3}{1} p^{1}q^{2}=3*0.2857*0.5102=0.4373\\\\\\P(x=2) = \dbinom{3}{2} p^{2}q^{1}=3*0.0816*0.7143=0.1749\\\\\\P(x=3) = \dbinom{3}{3} p^{3}q^{0}=1*0.0233*1=0.0233\\\\\\$

7 0

2 years ago

A vehicle 1 1/8 uses gallons of gasoline to travel 13 1/2 miles. At this rate, how many miles can the vehicle travel per gallon

alexgriva [62]

Answer: 12 miles per gallon

trust

mama

this

correct

3 0

3 years ago

Read 2 more answers

Find the surface area for the following figures PLease help

Yakvenalex [24]

Answer:

RP=190

TP=136

C=1,099

Step-by-step explanation:

RECTANGULAR PRISM, the equation is A=2(wl+hl+hw)

A=5*5+7*5+7*5

A=25+35+35

A=95*2

A=190

TRIANGULAR PRISM, the equation is A=1/2bh for the top and bottom triangles and A=lw for the sides

A=1/2(6)(4)

A=1/2*24

A=12

A=7*5

A=35

A=7*6

A=42

12*2=24

35*2=70

24+70+42=136

CYLINDER, the equation is A=2πrh+2πr2

A=2*3.14(7)(18)+2*3.14(7)^2

A=6.28*126+6.28*49

A=791.28+307.72

A=1,099

5 0

2 years ago

2. A statistics student plans to use a TI-84 Plus calculator on her final exam. From past experience, she estimates that there i

Anarel [89]

Answer:

P(≥1 working) = 0.9936
She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.

Step-by-step explanation:

1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...

... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936

2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.

If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.

This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.

_____

My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)

7 0

3 years ago