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Ad libitum [116K]

3 years ago

3

joaquin can send 250 text each month so far this month he has sent 141 text message let T represent the number of text messages

joaquin can send during the rest of the month. Write an inequality to model the situation. Solve the inequality for t.

1 answer:

Naddik [55]3 years ago

7 0

Answer:

109 texts the rest of the month

Step-by-step explanation:

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Find volume of rectangular prism length(4x+3) width (x-6) height (2x-1)

attashe74 [19]

Answer:

8x^3-46x^2-5x+18

Step-by-step explanation:

The volume of a rectangular prism is L*W*H where

L=length

W=width

H=height.

So we want to probably find the standard form of this multiplication because writing (4x+3)(x-6)(2x-1) is too easy.

Let's multiply (4x+3) and (x-6), then take that result and multiply it to (2x-1).

(4x+3)(x-6)

I'm going to use FOIL here.

First: 4x(x)=4x^2

Outer: 4x(-6)=-24x

Inner: 3(x)=3x

Last: 3(-6)=-18

---------------------------Add.

4x^2-21x-18

So we now have to multiply (4x^2-21x-18) and (2x-1).

We will not be able to use FOIL here because we are not doing a binomial times a binomial.

We can still use distributive property though.

(4x^2-21x-18)(2x-1)

=

4x^2(2x-1)-21x(2x-1)-18(2x-1)

=

8x^3-4x^2-42x^2+21x-36x+18

Now the like terms are actually already paired up we just need to combine them:

8x^3-46x^2-5x+18

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4 years ago

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Square root for the number 25/4

svp [43]

5/2 ....................................

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4 years ago

Suppose a certain study reported that 27.7% of high school students smoke.

Norma-Jean [14]

<u>Given</u>:

Probability of student smoke,

P = 27.7%

= 0.277

Number of students,

n = 632

$q = 1-p$

$=1-0.277$

$=0.723$

(i)

Here,

Number of students (n) = 60

then,

⇒ $n_P=60\times 0.277$

$=16.62$

⇒ $n_q=60\times 0.723$

$=43.38$

We can see that $n_P > 10$ and $n_q>10$ so the normal approximation condition are met.

Now,

$\mu = n_P= 16.62$

$\sigma = \sqrt{n_{Pq}}$

$= \sqrt{60\times 0.277\times 0.723}$

$=3.9664$

Now,

⇒ $P(X$

$=P(Z_{18.5})$

The Z-score is:

= $\frac{18.5-16.62}{3.4664}$

= $0.5423$

hence,

The probability will be:

⇒ $P(Z_{18.5}) = 0.7062$

or,

⇒ $P(Z$

(ii)

Here,

Number of students (n) = 75

$\mu = n_P = 75\times 0.277$

$=20.775$

$\sigma = \sqrt{n_{Pq}}$

$=\sqrt{75\times 0.277\times 0.723}$

$=3.8756$

Now,

⇒ $P(X>17) = P(X> 17.5)$

$=1-P(X \leq 17.5)$

$=1-P(Z_{17.5})$

The Z-score is:

= $\frac{17.5-20.775}{3.8756}$

= $-0.9740$

then, $P(Z_{17.5}) = 0.165$

hence,

The probability will be:

⇒ $P(X>17) = 1-0.165$

$=0.835$

Learn more about probability here:

brainly.com/question/9825651

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What are the solutions to |5x+2|=8

netineya [11]

Decimal Form:

x=2,−1.2

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4 years ago

Solve for x. Show each step of the solution. 4.5(4 − x ) + 36 = 202 − 2.5(3x + 28)

Marrrta [24]

Answer:

sgsrgrsdg

Step-by-step explanation:

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