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klio [65]
3 years ago
13

Tiffiny ate 1 slice of cake. Omer ate 2 slices. If Tiffiny ate 1/5 of the cake and all the slices were the same size, what fract

ion of the cake remained after Tiffiny and Omer had eaten?
Mathematics
2 answers:
Studentka2010 [4]3 years ago
5 0
If the total of the slices is one cake, and each slice is '<em>s</em>' in size, then you could model an equation like this:
Total (one cake) = 1 slice + 2 slices + 1/5 slices + remaining cake, or
c = 1s + 2s + \frac{1}{5}s + R\\&#10;c=2\frac{1}{5}s+R\\&#10;R=2\frac{1}{5}s -c
Now since I'm sure this question is incomplete, plug in whatever values you need to get the answer.
Nutka1998 [239]3 years ago
3 0
2/5 of the cake are left. Tiffany ate 1/5 and Omer ate 2x the amount, which is 2/5.
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A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co
Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let \mu = <u><em>mean weight of coffee in its containers.</em></u>

SO, Null Hypothesis, H_0 : \mu \geq 32 ounces     {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, H_A : \mu < 32 ounces      {means that the mean weight of coffee in its containers is less than 32 ounces}

The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;

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

where, \bar X = sample mean weight = 31.8 ounces

             s = sample standard deviation = 0.48 ounces

             n = sample of containers = 15

So, <u><em>the test statistics</em></u>  =  \frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }  ~ t_1_4  

                                      =  -1.614

The value of t test statistics is -1.614.

<u>Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

8 0
4 years ago
Help i will give brainlist
Arisa [49]

Answer:

3. 2/10 4/20 3/15

4. 12/1 48/4 36/3

Step-by-step explanation:

3 0
3 years ago
Complete the steps to solve the polynomial equation x3 – 21x = –20. According to the rational root theorem, which number is a po
jeka94

Answer:

Zeroes : 1, 4 and -5.

Potential roots: \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.

Step-by-step explanation:

The given equation is

x^3-21x=-20

It can be written as

x^3+0x^2-21x+20=0

Splitting the middle terms, we get

x^3-x^2+x^2-x-20x+20=0

x^2(x-1)+x(x-1)-20(x-1)=0

(x-1)(x^2+x-20)=0

Splitting the middle terms, we get

(x-1)(x^2+5x-4x-20)=0

(x-1)(x(x+5)-4(x+5))=0

(x-1)(x+5)(x-4)=0

Using zero product property, we get

x-1=0\Rightarrow x=1

x-4=0\Rightarrow x=4

x+5=0\Rightarrow x=-5

Therefore, the zeroes of the equation are 1, 4 and -5.

According to rational root theorem, the potential root of the polynomial are

x=\dfrac{\text{Factor of constant}}{\text{Factor of leading coefficient}}

Constant = 20

Factors of constant ±1, ±2, ±4, ±5, ±10, ±20.

Leading coefficient= 1

Factors of leading coefficient ±1.

Therefore, the potential root of the polynomial are \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.

3 0
4 years ago
Estimate 13.23 - 2.570 by first rounding each number to the nearest tenth.
Temka [501]

Answer:

7

Step-by-step explanation:

13.23 = 10

2.570= 3

10-3=7

6 0
3 years ago
Read 2 more answers
A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of
Gwar [14]

\bold{\huge{\underline{ Solution}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • A marker in the center of the fairway is 150 yards away from the centre of the green
  • While standing on the marker and facing the green, the golfer turns 100° towards his ball
  • Then he peces off 30 yards to his ball

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>distance </u><u>between </u><u>the </u><u>golf </u><u>ball </u><u>and </u><u>the </u><u>center </u><u>of </u><u>the </u><u>green </u><u>.</u>

<h3><u>Let's </u><u> </u><u>Begin </u><u>:</u><u>-</u></h3>

Let assume that the distance between the golf ball and central of green is x

<u>Here</u><u>, </u>

  • Distance between marker and centre of green is 150 yards
  • <u>That </u><u>is</u><u>, </u>Height = 150 yards
  • For facing the green , The golfer turns 100° towards his ball
  • <u>That </u><u>is</u><u>, </u>Angle = 100°
  • The golfer peces off 30 yards to his ball
  • <u>That </u><u>is</u><u>, </u>Base = 30 yards

<u>According </u><u>to </u><u>the </u><u>law </u><u>of </u><u>cosine </u><u>:</u><u>-</u>

\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}

  • Here, a = perpendicular height
  • b = base
  • c = hypotenuse
  • cos theta = Angle of cosine

<u>So</u><u>, </u><u> </u><u>For </u><u>Hypotenuse </u><u>law </u><u>of </u><u>cosine </u><u>will </u><u>be </u><u>:</u><u>-</u>

\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}

\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}

\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}

\sf{ x^{2} = 22500 + 900 + 156.6}

\sf{ x^{2} = 23556.6}

\bold{ x = 153.48\: yards }

Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards

5 0
3 years ago
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