The solutions it suppose to equal W= -2/2, 0

HELP ASAP WILL GIVE FIRST PERSON BRAINLIEST

kaheart [24]

30 * 1/5 = 6 so 6 have a cell phone

1/2 * 6 = 3 so 3 have a cellphone and social media

195 divide by 2 1/2

195 divide 5/2 (copy dot flip)

195 * 2/5

78 loads of laundry

19.99 divide by 78

.256

.26

he is paying 26 cents for each load

4 0

3 years ago

An outside thermometer reads -4 degrees at 7:00 a.m. By 2:30 p.m. the temperature outside has increased by 11 degrees. At 2:30 p

Gnesinka [82]

Answer:

-169 degrees

Step-by-step explanation:

It doesn't make much sense... but let's answer the question anyway.

First task is to determine what's the temperature outside at 2:30 PM.

We know that it's 11 degrees cooler than it was at 7 AM, so -4 - 11 = -15 degrees.

Abram's house is 4 degrees less than 11 times the outside temperature, so we can express that as:

House = 11 x - 4 - where x is the temperature outside.

If we replace the numbers we get: 11 * (-15) - 4 = -165 - 4 = -169 degrees

Abram's house is an extremely cold freezer!

3 0

3 years ago

Read 2 more answers

Find f'(x) if f(x) = (2/(x^1/3)) + 3 cos x + x^pi

galben [10]

The derivative of $f$ is

$f'(x)=\left(2x^{-1/3}+3\cos x+x^\pi\right)'$

$f'(x)=\left(2x^{-1/3}\right)'+(3\cos x)'+\left(x^\pi\right)'$

By the power rule,

$f'(x)=-\dfrac23x^{-4/3}+(3\cos x)'+\pi x^{\pi-1}$

The derivative of $\cos x$ is $-\sin x$ :

$f'(x)=-\dfrac23x^{-4/3}-3\sin x+\pi x^{\pi-1}$

3 0

3 years ago

A cylinder is placed within a cube. The diameter of the cylinder is equal to the length of each side of the cube. What is the vo

pochemuha

The volume of a shape is the amount of space in it.

The volume outside the cylinder, but inside the cube is 214.6 cubic inches

The given parameter is:

$l = 10$ ---- length of the cube

So, the volume of the cube is:

$V_1 = l^3$

$V_1 = 10^3$

$V_1 = 1000$

The volume of the cylinder is:

$V_2 = \pi r^2h$

The diameter (d) of the cylinder equals the length of the cube.

This means that:

$d = l$

$d = 10$

Divide by 2 to calculate the radius of the cylinder

$r = \frac{10}{2}$

$r = 5$

Also, the height of the cube is::

$h =l$

$h =10$

So, we have:

$V_2 = \pi r^2h$

$V_2 = \pi \times 5^2 \times 10$

$V_2 = \pi \times 25 \times 10$

$V_2 = \pi \times 250$

$V_2 = 785.4$

So, the volume (V) outside the cylinder is:

$V = V_1 - V_2$

$V = 1000 - 785.4$

$V = 214.6$

Hence, the volume outside the cylinder is 214.6 cubic inches

Read more about volumes at:

brainly.com/question/15861918

5 0

2 years ago

Suppose we have two thermometers. One thermometer is very precise but is delicate and heavy (X). We have another thermometer tha

Temka [501]

Answer:

a)H0: β1= 1 and Ha: β1 ≠1

b) The test statistic is____.t= b- β/ sb

c) The p-value is____.0.999144

(d) Therefore, we can conclude that:_____. there is not enough evidence to reject the null hypothesis.

2) The data provides no evidence at the 0.1 significance level that these thermometers are not consistent.

Step-by-step explanation:

The null and alternate hypotheses are

H0: β1= 1 and Ha: β1 ≠1

The significance level is alpha= 0.1 but ∝/2 =0.1/2= 0.05 for two tailed test.

The critical region at t ∝/2 (29)= t ≤ - 1.699 , t ≥ 1.699

The test statistic is

t= b- β/ sb

which has t distribution with υ= 31-2= 29 degrees of freedom.

Calculations

Ŷ = a +bX

b = SPxy /SS x= Σ(xi-x`)(yi-y`)/Σ(xi-x`)²

b = 39671.6/39680= 0.9998

a = y` - bx`

x`= 60

y`= 59.989

a = 59.989 -0.9998*60 = 0.001734

Syx²= Σ( yi -y`)²/ n-2= 39663.3857/ 29=1367.68965

Syx= 36.9822

Sb= Syx/ √∑ (x-x`)²

Sb= 36.9823/ √39680

Sb= 36.9823/ 199.984

Sb= 0.18493

x-x` y-y` (x-x`)² (x-x`)(y-y`)

-60 -59.969 3600 3598.1419

-56 -55.999 3136 3135.9458

-52 -52.079 2704 2708.1097

-48 -47.959 2304 2302.0335

-44 -43.899 1936 1931.5574

-40 -39.989 1600 1599.5613

-36 -36.009 1296 1296.3252

-32 -31.899 1024 1020.769

-28 -28.049 784 785.3729

-24 -23.959 576 575.0168

-20 -19.989 400 399.7806

-16 -15.939 256 255.0245

-12 -12.039 144 144.4684

-8 -7.989 64 63.9123

-4 -4.119 16 16.4761

0 -0.08903 0 0

4 3.921 16 15.6839

8 7.961 64 63.6877

12 12.121 144 145.4516

16 16.031 256 256.4955

20 20.021 400 400.4194

24 24.111 576 578.6632

28 28.071 784 785.9871

32 31.751 1024 1016.031

36 36.031 1296 1297.1148

40 39.961 1600 1598.4387

44 43.881 1936 1930.7626

48 48.021 2304 2305.0065

52 52.001 2704 2704.0503

56 56.051 3136 3138.8542

60 60.041 3600 3602.4581

∑0 0 39680 (SSx) 39671.6 (SPxy)

Putting the values

The test statistic is

t= b- β/ sb

t= 0.9998-1/ 0.18493 ( β=1 given)

t=-0.0010815

Since the calculated t=-0.0010815 does not lie in the critical region t ∝/2 (29)= t ≤ - 1.699 , t ≥ 1.699 we conclude that these thermometers seem to be measure temperatures.

The p-value is ≈ 0.999144

there is not enough evidence to reject the null hypothesis.

7 0

3 years ago

The solutions it suppose to equal W= -2/2, 0​

The solutions it suppose to equal W= -2/2, 0