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3 years ago

Will the sum of 56+ (-56) be Positive negative or zero

Mathematics

2 answers:

MrRissso [65]3 years ago

8 0

0 is the answer! U can type it in the calculator

d1i1m1o1n [39]3 years ago

5 0

Answer:

zero

Step-by-step explanation:

56+(-56),thus

the negative overcomes the positive,

56-56

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Mrs Martin deposits $2,350 in a money market account. Her account earns an interest rate 18% if she does not deposit or withdraw

VARVARA [1.3K]

Answer: $2,561.50
(I = A - P = $211.50)

Equation:
A = P(1 + rt)
Calculation:
First, converting R percent to r a decimal
r = R/100 = 18%/100 = 0.18 per year.

Solving our equation:
A = 2350(1 + (0.18 × 0.5)) = 2561.5
A = $2,561.50

The total amount accrued, principal plus interest, from simple interest on a principal of $2,350.00 at a rate of 18% per year for 0.5 years is $2,561.50.

4 0

3 years ago

Ryan drove 5 3/4 miles to the bank , 6 1/2 miles to the post office and 4 1/2 miles to the park. What was the total distance Rya

OLga [1]

Answer:

16 3/4 miles

Step-by-step explanation:

So the goal of this problem is to get Ryan's total driving distance which was made up of:

A 5 3/4 mile trip

A 6 1/2 mile trip

A 4 1/2 mile trip

Start by adding the last two since they have the same denominator

6 1/2 + 4 1/2 = 10 2/2 = 11

11 miles plus 5 3/4 miles

11 + 5 3/4 = 16 3/4

He drove at total of 16 3/4 miles

6 0

3 years ago

What kind of relationship is expressed below? <br> 5x-9>12

omeli [17]

Answer:

Today: Tuesday, 12th January 2021

21.16 PM

$\tt 5x - 9 > 12$

$\tt 5x > 12 + 9$

$\tt 5x > 21$

${\orange{\boxed{ \red{\tt x = \frac{21}{5} }}}}$

8 0

3 years ago

A study reveals that 65% of homeowners in Omaha own a lawn mower. To test this claim, you form a sample of 500 homeowners in Oma

kotegsom [21]

Answer:

There is no enough evidence to reject the claim using a level of significance of 5%

Step-by-step explanation:

Let p be the true proportion of homeowners in Omaha who own a lawnmower. We want to test

$H_{0}: p = p_{0} = 0.65$ vs $H_{1}: p\neq 0.65$ (two-tailed alternative).

We have a large sample of size n = 500 and a point estimate for p is $\hat{p}=0.68$ . The test statistic is given by

$Z=\frac{\hat{p}-p_{0}}{\sqrt{p_{0}(1-p_{0})/n}}=\frac{0.68-0.65}{\sqrt{0.65(1-0.65)/500}}$ where $\sqrt{0.65(1-0.65)/500}$ is the standard error of $\hat{p}$ when $H_{0}$ is true. The observed value is $z_{0}$ = 1.4064 which comes from a standard normal distribution. Because we want to use a level of significance of 5%, our rejection region is RR={z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles respectively. Because 1.4064 does not belong to RR, we fail to reject the null hypothesis.