The price of a ring was increased by 8% to £486. What was the price before the increase?

Tpy6a [65]

Sin(19)

Use your calculator.

sin(19) = 0.3255681545

6 0

2 years ago

PLEASE HELP! The avenues in a particular city run north to south and are numbered consecutively with 1st Avenue at the western b

AveGali [126]

Answer: D

Step-by-step explanation:

Just did it

4 0

2 years ago

Read 2 more answers

Marisol is trying to decide which job to take. She is offered three jobs. The store manager has a salary of $32,500. The

Anon25 [30]

Answer:

I think it is the department manager

8 0

2 years ago

On Monday, the closing value of a share of stock in Company ABC, was 74.01. On Tuesday, it closed at 73.67, and on Wednesday, it

Elenna [48]

Answer:

$957.71$

Step-by-step explanation:

Given: On Monday, the closing value of a share of stock in Company ABC, was $74.01$ . On Tuesday, it closed at $73.67$ , and on Wednesday, it closed at $74.32$ . On Thursday, the change in stock value was $-1.48$ . And, on Friday, Company ABC's stock gained $0.89$ points.

To Find: One shareholder owns $13$ shares of Company ABC's stock. What was the total value of their stock in this company at the close of Tuesday.

Solution:

Closing value of Shares of company ABC on Monday $=74.01$

Closing value of Shares of company ABC on Tuesday $=73.67$

Closing value of Shares of company ABC on Wednesday $=74.32$

on Thursday change in stock value was $-1.48$

Closing value of Shares of company ABC on Thursday $=72.84$

on Friday comapny's stock gained $0.89$ points

Closing value of Shares of company ABC on Friday $=73.73$

As one shareholder owns $13$ shares of company

Total value of shares on Tuesday is

$=\text{number of shares}\times\text{closing stock value on tuesday}$

$=13\times73.67$

$=957.71$

Hence total value of shareholder's share on close of Tuesday is $957.71$

6 0

2 years ago

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A large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times

WARRIOR [948]

Answer:

We conclude that the new procedure will not decrease the population mean amount of time required to produce the part.

Step-by-step explanation:

We are given that a large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours.

A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours

Let $\mu$ = population mean amount of time required to produce an electrical part using new procedure

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 45 hours {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}

Alternate Hypothesis, $H_a$ : $\mu$ < 45 hours {means that the new procedure will decrease the population mean amount of time required to produce the part}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\mu$ = sample mean amount of time = 43.118 hours

s = sample standard deviation = 5.5 hours

n = sample of parts = 25

So, test statistics = $\frac{43.118-45}{{\frac{5.5}{\sqrt{25} } } }$ ~ $t_2_4$

= -1.711

Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.

4 0

2 years ago