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guapka [62]
3 years ago
15

How many almonds in the cup? (1/4 cup) Help me​

Mathematics
1 answer:
IgorLugansk [536]3 years ago
7 0

Answer:

The amount of almonds in the fourth cup are 40

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Statistics help please!
Anastasy [175]
The answer would be $9
7 0
3 years ago
Solve inequality in interval notation Y^2>-5y-9
nata0808 [166]
It’s going to be negative infinitely, infinitely.
4 0
2 years ago
Please help me:(I’ll give you the brainliest
Savatey [412]

Answer:

The range of values for x is; 5 < x < 29

Step-by-step explanation:

The given parameters are;

\overline {AB} = 15

\overline {AD} = 18

\overline {BC} = \overline {CD} Given

\overline {AC} = \overline {AC} by reflexive property

∠BCA = 2·x - 10

∠DCA = 48°

Since 15 < 18, given the common sides of the tringle ΔABC and triangle ΔADC, angle ∡(2·x - 10)° < 48°

Therefore;

2x - 10 < 48

2·x < 48 + 10

∴ x < 29

Also, given that 2·x - 10 is real, 0 < 2·x - 10

10 < 2·x

10/2 < x

5 < x

Therefore, an acceptable range is 5 < x < 29

4 0
2 years ago
A student wants to report on the number of movies her friends watch each week. The collected data are below: 2 14 1 2 0 1 0 2 Wh
Sunny_sXe [5.5K]

The student's friend watch 1.5 movies every week.

Data provided to us {2, 14, 1, 2, 0, 1, 0, 2}.

Number of data values = 8

Sorted or rearranged data {0, 0, 1, 1, 2, 2, 2, 14}

Median of the data provided;

Median is the mid point value of the data provided to us, but as in this case there are 8 data provided (even number of data), we will take the average of the 2 mid values of the data.

Mid Values = 1 and 2

\bold{Median=\dfrac{1+2}{2} = 1.5}

Mean of the data provided;

Mean=\dfrac{Sum\ of\ all\ data\ values}{Number\ of\ data\ values} \\\\Mean= \dfrac{(0+0+1+1+2+2+2+14)}{8}\\\\Mean=2.75

As we know, The data values provided are 0, 1 and 2, and there is 14 as well in the data which is the maximum value in the data provided to us but the value is far from the other data values provided to us. Therefore, the data value 14 is considered to be an outlier.

An outlier is the maximum value that increases the mean, which makes it larger than it should be.

If we were to compute the mean without the outlier,

Mean=\dfrac{Sum\ of\ all\ data\ values}{Number\ of\ data\ values} \\\\Mean= \dfrac{(0+0+1+1+2+2+2)}{7}\\\\Mean=1.1428

Now, we have understood that the mean is sensitive to outlier can be easily manipulated. thus, mean is not the right option to go for in this case.

Therefore, the median value of 1.5 is the right option.

Hence, the student's friend watch 1.5 movies every week.

To know more visit:

brainly.com/question/1363341

3 0
2 years ago
Read 2 more answers
Please help! Consider the given circle with the shaded sector xy and central angle 225. The circumference is 30 units
OleMash [197]

\textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=30\pi \end{cases}\implies 30\pi =2\pi r\implies \cfrac{30\pi }{2\pi }=r\implies \boxed{15=r} \\\\[-0.35em] ~\dotfill

\textit{arc's length}\\\\ s = \cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=15\\ \theta =225 \end{cases}\implies \widehat{XY}=\cfrac{(225)\pi (15)}{180}\implies \boxed{\widehat{XY}\approx 58.90} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of a circle}\\\\ A=\pi r^2\qquad \qquad A=\pi (15)^2\implies A=225\pi \implies \boxed{A\approx 706.86}

6 0
2 years ago
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