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3 years ago

Pls help me if u do i will give u the best answer (well of course)

Mathematics

2 answers:

jolli1 [7]3 years ago

4 0

Do we have to solve for x?

laila [671]3 years ago

3 0

Let's solve your equation step-by-step.
14(3x−16)+14x=513
Step 1: Simplify both sides of the equation.
x−4=163
Step 2: Add 4 to both sides.
x−4+4=163+4x=283
Answer:
x=913

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Use the drawing tool(s) to form the correct answer on the provided graph.

Kitty [74]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$y> \frac{1}{4}x+6$ ------>inequality A

The solution of the inequality A is the shaded area above the dashed line $y=\frac{1}{4}x+6$

The y-intercept of the dashed line is (0,6)

The x-intercept of the dashed line is (-24,0)

The slope of the dashed line is positive m=1/4

$y> 2x-1$ ------>inequality B

The solution of the inequality B is the shaded area above the dashed line $y=2x-1$

The y-intercept of the dashed line is (0,-1)

The x-intercept of the dashed line is (0.5,0)

The slope of the dashed line is positive m=2

The solution of the system of inequalities is the shaded area between the two dashed lines

using a graphing tool

see the attached figure

4 0

3 years ago

Seven less than twice a number w is 15.What is the number?

slega [8]

Answer:

Step-by-step explanation:

seven less than twice a number equals 19. = 13. Answer.

3 0

3 years ago

Read 2 more answers

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

To make a specific orange color, a painter uses 14 of a can of red paint and 12 of a can of yellow paint. Fill in each statement

Otrada [13]

Answer: For every 1 can of red paint, the number of yellow paints used by the painter is $\frac{6}{7}$ and

There are approximately 29 cans of yellow paints for 34 cans of red paints.

Step-by-step explanation:

Since we have given that

Number of cans of red paint = 14

Number of cans of yellow paint = 12

According to question, we have to find that for every 1 can of red paint the painter uses what number of yellow paints;

Since the ratio of red paint to yellow paint is given by

$14:12\\\\7:6\\\\1:\frac{6}{7}$

So, for every 1 can of red paint, the number of yellow paints used by the painter is $\frac{6}{7}$

Similarly,

If Number of can of red paint is used = 34

So, Number of cans of yellow paint will be

$\frac{6}{7}\times 34\\\\=\frac{6\times 34}{7}\\\\=29\ approx.$

Hence, there are approximately 29 cans of yellow paints for 34 cans of red paints.

3 0

3 years ago

a tea company surveyed shoppers at two different malls. shoppers were asked whether they prefer sweetened or unsweetened tea. at

FromTheMoon [43]

First, let's create the ratios of sweetened to unsweetened. To do this, you would have to subtract the number of people that preferred unsweetened from the total.
Westside mall: 15:30 or $\frac{15}{30}$
Eastside mall: 13:26 or $\frac{13}{26}$

Now, you would divide to both ratios.
15 ÷ 30 = 0.5
13 ÷ 26 = 0.5

They have the same sum, meaning, they are the same.

Westside Mall shoppers are just as likely to prefer unsweetened tea as Eastside Mall shoppers.

I hope this helps!

8 0

3 years ago