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1 year ago

What integer describes 50 feet below sea level?

Mathematics

2 answers:

alexdok [17]1 year ago

8 0

-50 since an integer is the opposite of a whole number positive or negative

ANEK [815]1 year ago

7 0

The answer is -50 because sea level is 0 and you are taking away 50 feet

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Which are examples of unit rates? Check all that apply.

mylen [45]

I believe the answer is $100 for 5 tickets. I hope this helps!

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2 years ago

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Mr Henderson has 2 bouncy ball vending machines , he buys one bag of the 27 millimeter balls and one bag of the 40 millimeter ba

weeeeeb [17]

I think 32
But I'm not 100% sure

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2 years ago

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The formula for the future value V (in dollars) of an investment earning simple interest is V=p+prt, where p (in dollars) is the

zhannawk [14.2K]

V=p+prt

now we solve for P

V=P (1+rt)

Divide both sides by (1+rt)

P=V÷(1+rt)...answer

P=V÷(1+rt)

P=3,000÷(1+0.06×5)

P=2,307.69

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2 years ago

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of

Thepotemich [5.8K]

<h2><u>Answer with explanation</u>:</h2>

As per given , we have

sample size : n= 65

degree of freedom : df=n-1=64

sample mean : $\overline{x}=19.5$

sample standard deviation : s= 5.2

Since , the population standard deviation is not given , so we apply t-test.

Significance level for 90% confidence : $\alpha=1-0.90=0.10$

t-critical value for significance level 0.10 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.6690$

Formula for Confidence interval :

$\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}$

Then , 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.669)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.076$

$(19.5-1.076,\ 19.5+1.076)=(18.424,\ 20.576)$

∴ 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.424, 20.576)

Significance level for 95% confidence : $\alpha=1-0.95=0.05$

t-critical value for significance level 0.05 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.9977$

Then , 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.9977)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.288$

$(19.5-1.288,\ 19.5+1.288)=(18.212,\ 20.788)$

∴ 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.212, 20.788)

4 0

2 years ago

Kara starts tying a baby quilt at the local community center. Each quilt takes 200 ties to complete. Ten minutes later Julie joi

Tju [1.3M]

Answer:

a) $K(t)=6\frac{tie}{min}(t)$

b) $J(t)=11\frac{tie}{min}(t)$

c) $T_{f}(t)=K(t+10)+J(t)$ where $T_{f}(t)$ = number ties finished per unit time

d) t≅8.24min

e) K=109tie

J=91tie

f) K≅16.6min

J≅9.09min

Step-by-step explanation:

Data

$Kara_{(K)}=6\frac{tie}{min}\\Julie(J)=11\frac{tie}{min}$

Kara begins ten minutes before julie, so when julie joins kara she has 60 ties

t=10 $k=6\frac{tie}{min}*10min=60tie$

a) $K(t)=6\frac{tie}{min}(t)$

b) $J(t)=11\frac{tie}{min}(t)$

c) $T_{f}(t)=K(t+10)+J(t)$ where $T_{f}(t)$ = number ties finished per unit time

d) if kara already has 60 ties and makes 6 per minute and Julie makes 11 per minute then per minute they makes 17 ties, so one minute per 17 ties for 200 ties ¿how long?. $t=\frac{(200-60)tie*min}{17tie}$ ≅8.24min

e) K=60tie+6tie/min*8.24min=109tie

J=11tie/min*8.24min=91tie

f) K=100tie*min/6tie≅16.6min

J=100tie*min/11tie≅9.09min

Note: the amount of ties were rounded to the nearest decimal

6 0

1 year ago