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3 years ago

WILL BE BRANLIEST!!!

Mathematics

2 answers:

stiks02 [169]3 years ago

8 0

Answer:

See explanation

Step-by-step explanation:

A. Solution set is in the attachment (dark green shading in the middle)

B. Yes, 2(5) + 3(1) ≤ 15; 13 is less than or equal to 15

5 + 1 > 3, 6 is greater than 3

C. (4, 2) means that he buys 4 sandwiches and 2 hot lunches

marishachu [46]3 years ago

3 0

Tfgha hey dggdff F ffvdvhuhdghj

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On a six-question multiple-choice test there are five possible answers for each question, of which one is correct and four are i

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

given that on a six-question multiple-choice test there are five possible answers for each question, of which one is correct and four are incorrect

By mere guessing probability for choosing correct answer = p = 0.2

also each question is independent of the other

Hence if X is the number of correct questions then

X is binomial with p = 0.2 and n = 6

a) being correct on three questions,

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3 years ago

A car travels 0.75 miles every minute

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Step-by-step explanation:

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What is the answer to the equation <br> (4 x 6) - 25 + 1

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A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

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3 years ago

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V125BC [204]

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