All categories

3 years ago

mmmbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

Mathematics

2 answers:

Iteru [2.4K]3 years ago

6 0

Answer:

mmmbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

Step-by-step explanation:

yuh get into ittttt

artcher [175]3 years ago

3 0

dang im gettin hell.a points today fhfjfjfjfjfjfjfkfkfkfjfjfjfjfjfjfjfjfjffjfjfjjfjfjfjjfjfjfjfjfjfjfjfjffjjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjfjffjjfjfjfjfjfjfjfjfjjfjfjfjfjfjfjfjfjfjfjfjjfj

You might be interested in

On saturday, a minor league baseball team gave away baseball cards to each person entering the stadium. one group received 28 ba

Tanzania [10]

29 card a piece! Hope it helps!

4 0

3 years ago

A particle moves so that r(t) = ati + b sin atj. Show that the magnitude of the acceleration of the particle is proportional to

ValentinkaMS [17]

Answer:

Step-by-step explanation:

Given

Position of particle is $r(t)=at\hat{i}+b\sin (at)\hat{j}$

i.e. distance from x axis is $b\sin (at)---1$

Distance from y axis at

velocity is given by $v=\frac{\mathrm{d} r}{\mathrm{d} t}$

$v=a\hat{i}+ba\cos (at)\hat{j}$

Similarly acceleration is given by

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=0\hat{i}-a^2b\sin (at)\hat{j}$

Magnitude of acceleration is $=\sqrt{(-a^2b\sin (at))^2}$

$=a^2b\sin (at)----2$

From 1 and 2 we can see that

Magnitude of acceleration is proportional to distance from x axis

$a\propto distance\ from\ x-axis$

5 0

3 years ago

Read 2 more answers

A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air

Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : p = 0.6

H₁ : p = 0.6, this explains the acceptance region as;

p° ≤ $\frac{315}{500}$ =0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

= P (p°>0.63 | p=0.6)

where p° is represented as pI in the subsequent calculated steps below

= P $[\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]$

= P $[\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]$

= P $[Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]$

= P [Z > 1.37]

= 1 - P [Z ≤ 1.37]

= 1 - Ф (1.37)

= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

≅ 0.0853

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

= P [p° ≤ 0.63 | p = 0.75]

where p° is represented as pI in the subsequent calculated steps below

= P $[\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]$

= P $[\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]$

= P $[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]$

= P [Z ≤ -6.20]

= Ф (-6.20)

≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0

3 years ago

The distance is approximately

Norma-Jean [14]

Distance = the square root of ((x2- x1) ^ 2 + (y2 - y1) ^ 2)

−You can use this to find your answer

−:)))

−

5 0

3 years ago

Show all steps 2a=22

aivan3 [116]

Answer:

Hello! answer: a = 11

Step-by-step explanation:

2 × 11 = 22 So therefore a = 11 HOPE THAT HELPS!

3 0

3 years ago