All categories

2 years ago

(5x+30)=7x ?? Can anyone help me

Mathematics

2 answers:

trasher [3.6K]2 years ago

7 0

Answer:

Step-by-step explanation:

(5x+30)=7x Subtract 7x from both sides. 5x+30−7x=0 Combine 5x and −7x to get −2x. −2x+30=0 Subtract 30 from both sides. Anything subtracted from zero gives its negation. −2x=−30 Divide both sides by −2. x= −2 −30 Divide −30 by −2 to get 15. answer should be x=15

Eddi Din [679]2 years ago

5 0

Answer:

the answer is x=15.....

You might be interested in

If m∠JKM = 43, m∠MKL = (8 - 20), and m∠JKL = (10x - 11), find each measure.

Brut [27]

Correct Question: If m∠JKM = 43, m∠MKL = (8x - 20), and m∠JKL = (10x - 11), find each measure.

1. x = ?

2. m∠MKL = ?

3. m∠JKL = ?

Answer/Step-by-step explanation:

Given:

m<JKM = 43,

m<MKL = (8x - 20),

m<JKL = (10x - 11).

Required:

1. Value of x

2. m<MKL

3. m<JKL

Solution:

1. Value of x:

m<JKL = m<MKL + m<JKM (angle addition postulate)

Therefore:

$(10x - 11) = (8x - 20) + 43$

Solve for x

$10x - 11 = 8x - 20 + 43$

$10x - 11 = 8x + 23$

Subtract 8x from both sides

$10x - 11 - 8x = 8x + 23 - 8x$

$2x - 11 = 23$

Add 11 to both sides

$2x - 11 + 11 = 23 + 11$

$2x = 34$

Divide both sides by 2

$\frac{2x}{2} = \frac{34}{2}$

$x = 17$

2. m<MKL = 8x - 20

Plug in the value of x

m<MKL = 8(17) - 20 = 136 - 20 = 116°

3. m<JKL = 10x - 11

m<JKL = 10(17) - 11 = 170 - 11 = 159°

3 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

The circle x2+y2=36 is translated 5 Units left and 4 units up where is the center of the new circle after the translation

Andrej [43]

The general equation of a circle is given by:
(x-a)^2+(x-b)^2=r^2
where:
(a,b) is the center
r is the radius

given the equation:
x^2+y^2=36
it means that the equation is centered at (0,0) with radius of 6 units. Thus a translation of 5 units to the left and 4 units up, will change the new center to
(-5,4)
thus the equation will be:
(x+5)^2+(y-4)^2=36

Answer: (x+5)^2+(y-4)^2=36

6 0

3 years ago

Make 54/2 to a mixed numbers

Burka [1]

Answer: 27

Step-by-step explanation: Look below for an attached explanation.

Hope this helps!

4 0

3 years ago

The tree diagram represents an

Leviafan [203]

Answer:

0.35

Step-by-step explanation:

.5x.7

6 0

3 years ago