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3 years ago

8

The old time train museum is raising the price of tickets for train rides.Tickets used to cost $4.80 each. The new ticket price

will be 1.25 times the old ticket price.What will the new ticket price be?

2 answers:

kogti [31]3 years ago

6 0

It's 6 because you times 4.80 times 1.25 and it's equivalent to 6

SVETLANKA909090 [29]3 years ago

5 0

Only add $4.80 + $1.25

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Solve each equation for y (get into slope-intercept/y=mx + b form):

Gwar [14]

Answer:

Step-by-step explanation:

In order to do this we need to isolate y by performing the inverse operations on the other values like so...

a) 10x + 5y = 20 ... subtract 10x on both sides

5y = 20 - 10x ... divide both sides by 5

y = 4 - 2x ... we can move the 2x to the right to make it into y = mx + b

y = -2x + 4

b) 3x - 2y = 10 + 4x ... subtract 3x on both sides

-2y = 10 + x ... divide both sides by -2

y = -5 - 0.5x ... move -0.5 to the left so it matches y = mx + b

y = -0.5x - 5

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

(c) 90.53%

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

Please choose the answer that describes the scientific notation for 5,098,000.

avanturin [10]

Scientific notation is the representation of a number in a power of ten, with a sequence of digits placing the comma after the first digit, and the power or then that is mathematically consistent with the number. For example 400 is 4.00 * 10^2; 3100 is 3.100 * 10^4; 13421 = 1.3421 * 10 ^4. It is important that you have only one digit before the comma. So, 5,098,00 is 5.098 * 1,000,00 = 5.098 * 10^6. <span>The answer is the option c. 5.098 * 10^6.</span>

8 0

3 years ago

High grade steel consists of 85% iron and 15% magnese. Low grade steel consists of 67% iron and 33% mag ese. NASA orders 500 ton

Mamont248 [21]

Answer:

$361\dfrac{1}{9}$ tons of high steel and $138\dfrac{8}{9}$ tons of low steel

Step-by-step explanation:

Let x be the number of tons of high grade steel and y be the number of tons ow low grade steel needed.

In x tons of high grade steel there are

$0.85x$ tons of iron

$0.15x$ tons of magnese

In y tons of low grade steel there are

$0.67y$ tons of iron

$0.33y$ tons of magnese

NASA orders 500 tons of steel, so

$x+y=500$

and specifies that it must be in the proportion 80% iron and 20% magnese, so

$0.85x+0.67y=500\cdot 0.8\\ \\0.85x+0.67y=400$

From the first equation,

$x=500-y$

Substitute it into the second equation:

$0.85(500-y)+0.67y=400\\ \\425-0.85y+0.67y=400\\ \\-0.18y=-25\\ \\y=\dfrac{2,500}{18}=138\dfrac{8}{9}\ tons\\ \\x=361\dfrac{1}{9}\ tons$

7 0

3 years ago

Round to the nearest whole number. Simplify (5(12 - 2)2 + 7]/3=

mario62 [17]

(5(12-2)2+7)/3

(5(10)2+7)/3

(50(2)+7)/3

(100+7)/3

107/3

35.66666...

<h2><em>36</em></h2>

7 0

3 years ago