All categories

2 years ago

NEED HELP ASAP PLEASEE!!!!!!!!!

1 answer:

7 0

The statement 6.05 can be read as "six and five-hundredths" is true and 11.8 can be read as "eleven and eight tens" is false

<h3>Place value</h3>

6.05

6 = ones

0 = tenths

5 = Hundredth

6.05 = six and five-hundredths

11.8

1 = Tens

1 = Ones

8 = tenths

11.8 = eleven and eight tenths

Learn more about decimal number:

brainly.com/question/1827193

#SPJ1

You might be interested in

for each x-y table given,copy the table, find the pattern and fill in the missing entries. Then write the rule for the pattern i

mash [69]

Answer:

see the attached for the table and rules

Step-by-step explanation:

a) A graph of the given points shows they lie on the same line, one with a slope of 3 and a y-intercept of 2. Thus the rule is ...

y = 3x +2

see the attachment for table values

b) The ratios of given points are all the same: y/x = 5/2, so that is the constant of proportionality:

y = (5/2)x

see the attachment for table values

8 0

4 years ago

A box office sold 147,523 tickets for an auto race.

Marysya12 [62]

Answer:

78,799 children's tickets

Step-by-step explanation:

147,523 total tickets, 68,724 adult tickets

Subtract to find amount of children's tickets:

147,523-68,724 = 78,799

5 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

4 years ago

You make $11.29 per hour and worked 26 hours last week. They took out 5.2% for

sashaice [31]

Your answer would be 56

5 0

3 years ago

If Adam buys 5 snakes at 20% off and each snake costs s dollars, which expression is NOT equivalent to the situation stated?A)4s

olga2289 [7]

C) 5s - 0.2s is not equivalent. This would be paying for 5 snakes and subtracting 20% of the price of one snake, not 20% of the price of all 5 snakes.

4 0

3 years ago

Read 2 more answers