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3 years ago

Write a few sentences about different ways to show subtraction for a problem lie 32-15.

Mathematics

2 answers:

Alexxx [7]3 years ago

8 0

If you want to subtract
Put the bigger number on the top and then the smaller then subtract

32
- 15
= 17

musickatia [10]3 years ago

5 0

You can show subtraction for a problem like 32-15 by simply writing it out just like you did, horizontally: 32-15=. You can also show subtraction for 32-15 by stacking the numbers, with the larger number on top: 32
-15

Another way to show subtraction for 32-15 is by treating it as subtracting numbers with different signs: 32 is the same as +32-15. Or, you can switch the numbers around: -15+32, the answer will be the same, 17.

Hope that helped!

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Can anyone help thank you ggg

polet [3.4K]

I believe it is 1. Function 2. Not a function 3. Function 4. Not a function. I hope this helps! :)

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3 years ago

Read 2 more answers

How is the series 9 + 13+ 17+ ... + 149 represented in summation notation?

VLD [36.1K]

Notice that

13 - 9 = 4

17 - 13 = 4

so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :

149 = 9 + 4k

140 = 4k

k = 140/4

k = 35

This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.

Put another way, we have

$\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=0}^{35} (9 + 4k)$

but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.

$\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k-1=0}^{35+1} (9 + 4(k-1))$

$\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=1}^{36} (5 + 4k)$

7 0

3 years ago

Which of the following shapes have an area of 48 square feet? Select all that apply.

yKpoI14uk [10]

I think 2,3,4 or B,C,D would be correct.

The area of a triangle is (b•h)/2
And the area of a rectangle and parallelogram would just be b•h

1) 8(6) = 48/2 =24

2) 9(10 2/3) = 96/2 =48

3) 19.2(2.5) = 48

4) 16(3) = 48

Hope this is right and hope it helps. Have a good day

4 0

3 years ago

a train starting from rest moves with uniform acceleration of 5 m/s2 .find its velocity when it has travelled a distance of 1 km

Degger [83]

The velocity of a train starting from rest moves with a uniform acceleration of 5 m/s2 would be 100 m/s.

Acceleration = 5 m/s2

Distance, s = 1 km = 1000m

Intital velocity, u=0 m/s

<h3>What is the first equation of motion?</h3>

We know that, Newton's equation of motion,

s = ut+ 1/2 at^2

1000 = 0 + 1/2 x 5 x t^2

1000 = 5/2 x t^2

2000/ 5 = t^2

t = 20 s

Also,a = (v-u)/t

5 = v-0/20

100 = v

v = 100 m/s

Hence, the velocity of a train starting from rest moves with a uniform acceleration of 5 m/s2 would be 100 m/s.

Learn more about velocity here;

brainly.com/question/3491339

#SPJ1

7 0

2 years ago

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed

tatyana61 [14]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 8 minutes

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes

(b) The P-value of the test statistics is 0.0436.

(c) We conclude that the actual mean waiting time equals the standard at 0.05 significance level.

(d) 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Step-by-step explanation:

We are given that the length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes.

A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation of 3.2 minutes.

Let $\mu$ = actual mean waiting time.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 8 minutes {means that the actual mean waiting time does not differs from the standard}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes {means that the actual mean waiting time differs from the standard}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

$\sigma$ = population standard deviation = 3.2 minutes

n = sample of shoppers = 120

So, test statistics = $\frac{8.5-8}{\frac{3.2}{\sqrt{120} } }$

= 1.71

The value of t test statistics is 1.71.

(b) Now, the P-value of the test statistics is given by;

P-value = P(Z > 1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.9564 = 0.0436

(c) Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual mean waiting time equals the standard.

(d) Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

$\sigma$ = population standard deviation = 3.2 minutes

n = sample of shoppers = 120

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population proportion, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $8.5-1.96 \times{\frac{3.2}{\sqrt{120} } }$ , $8.5+1.96 \times{\frac{3.2}{\sqrt{120} } }$ ]

= [7.93 minutes , 9.07 minutes]

Therefore, 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Yes, it support our above conclusion.

3 0

3 years ago