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3 years ago

(-9)-[15divided by 5]^3•6

Mathematics

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

-171

Step-by-step explanation:

Applying PEMDAS (order of operations),

We open the parenthesis first;

-9 - ( $\frac{15}{5}$ )³ × 6

-9 - 3³ × 6

We work on multiplication next;

-9 - 27 × 6 = -9 - 162

finally we do subtraction to get;

-9 - 162 = -171

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Marina CMI [18]

Answer:

A) The amount of money deposited monthly

Step-by-step explanation:

y = 35x (slope) (amount deposited each month)+ 250 (y-intercept) (starting amount)

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2 years ago

478 cm x 427 cm = ? meters

vekshin1

478×427= 204,106cm
204,106cm - 2041.06m
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3 years ago

Lin ran 2 3/4 miles in 2/5 of an hour, Noah ran 8 2/3 miles in 4/3 of an hour, how far would Lin run in 1 hour?

Marizza181 [45]

Answer:

$6\frac{7}{8}\ miles$

Step-by-step explanation:

Given that Lil's distance in 2/5 hrs is 2 3/4hrs

-Let x be the distance ran in 1 hr.

#We equate and cross multiply to solve for x as follows:

$\frac{2}{5}\ hrs=2\frac{3}{4}\\1\ hr=x\\\\\therefore x=2\frac{3}{4}/\frac{2}{5}\\\\=6\frac{7}{8}\ miles$

Hence, Lil can run 6 7/8 miles in 1 hr.

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3 years ago

Find the area of the region in two ways. a. Using integration with respect to x. b. Using geometry. 9-x

Natali [406]

Answer: hello your question is incomplete attached below is the complete question

answer :

a) $\int\limits^3_0 {(10-x)-(x+4)} \, dx$ ( option D )

b) A = 1/2 (6)(3) ( option B )

c) Area of shaded region = 9

Step-by-step explanation:

<u>a) Using integration with respect to x </u>

Area = $\int\limits^7_4 {(y-4)} \, dy + \int\limits^a_7 {(10-y)} \, dy$ ( note a = 10 )

= y^2/2 - 4y |⁷₄ + 10y - y^2/2 |¹⁰₇

= 33/2 - 12 + 30 - 51/2 = 9

hence the best integral from the options attached is option D

$\int\limits^3_0 {(10-x)-(x+4)} \, dx$

= [ 10x - x^2 /2 - x^2/2 - 9x ] ³₀

= 30 - 9/2 - 9/2 - 12 = 9

<u>b) Using Geometry </u>

Area = 1/2 * base * height

= 1/2 * 6 * 3

= 9

7 0

3 years ago

The bad debt ratio for a financial institution is defined to be the dollar values of loans defaulted divided by the total dollar

Nimfa-mama [501]

Answer:

(a) NULL HYPOTHESIS, $H_0$ : $\mu \leq$ 3.5%

ALTERNATE HYPOTHESIS, $H_1$ : $\mu$ > 3.5%

(b) We conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Step-by-step explanation:

We are given that a random sample of seven Ohio banks is selected.The bad debt ratios for these banks are 7, 4, 6, 7, 5, 4, and 9%.The mean bad debt ratio for all federally insured banks is 3.5%.

We have to test the claim of Federal banking officials that the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

(a) Let, NULL HYPOTHESIS, $H_0$ : $\mu \leq$ 3.5% {means that the the mean bad debt ratio for Ohio banks is less than or equal to the mean for all federally insured banks}

ALTERNATE HYPOTHESIS, $H_1$ : $\mu$ > 3.5% {means that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks}

The test statistics that will be used here is One-sample t-test;

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

where, $\bar X$ = sample mean debt ratio of Ohio banks = 6%

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 1.83%

n = sample of banks = 7

So, test statistics = $\frac{6-3.5}{\frac{1.83}{\sqrt{7} } }$ ~ $t_6$

= 3.614

(b) Now, at 1% significance level t table gives critical value of 3.143. Since our test statistics is more than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Hence, Federal banking officials claim was correct.

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3 years ago