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3 years ago

PLZZZ I NEED HELPPPPPP 4x-5=-8 solve for x plz help and show work I really want to know how to do it

Mathematics

1 answer:

nignag [31]3 years ago

4 0

Answer:

x= 3.25

Step-by-step explanation:

Lets start off by writing the equation: $4x-5=-8$

next you subtract 5 from both sides

then you will get: $4x=-13$

next you will divide 4 from both sides to isolate x and what is equal to x

then you get your answer :)

$4x-5=-8$

$4x-5=-8\\ -5 -5$

$4x=-13$

$\frac{4x=-13}{4}$

$x = -3.25$

i hope this helps luv :)

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Figure below shows two triangles EFG and KLM

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Answer:

last option

Step-by-step explanation:

To prove that ΔEFG is also a right triangle, you must prove that KL = EF so that in ΔKLM c² = a² + b² which would make ΔEFG a right triangle.

3 0

3 years ago

(4x + 8x - 5) + (3x2 + 6x - 7)

ch4aika [34]

Answer:

7x^2 + 14x - 12

Step-by-step explanation:

Add like terms:

(4x + 8x - 5) + (3x2 + 6x - 7) becomes:

4x^2 + 8x - 5 or (grouping like terms in columns):

+ 3x^2 + 6x - 7

---------------------

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5 0

3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago

A cookie factory monitored the number of broken cookies per pack yesterday.

trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

$s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } }$ $\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }$

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

8 0

3 years ago