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Gnom [1K]
3 years ago
15

1. Here are the driving scores of 40 new drivers. Make a relative frequency table for their scores:

Mathematics
1 answer:
Kobotan [32]3 years ago
8 0

From the question we are told that:

Sample size n=40

Generally Number of class (K) is Mathematically given as

2^k \geq 40

Therefore

2^k \geq 2^6

k=6

Generally with class interval of k

Class Width is  mathematically given as

W=\frac{highest\ score-lowest\ score}{k}

W=\frac{97-42}{6}

W=10

The complete Table is attached below

2)

The complete Image of the Histogram is attached below

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At parks elementary hot sandwiches are served on Mondays and Thursdays tacos on Tuesdays and salads on Wednesday’s if pizza was
Dimas [21]

Answer: Friday


Step-by-step explanation: If you had hot sandwiches, tacos, and salad on every day of the week except for Friday, then pizza would be served on Friday. Hope this helped!


4 0
3 years ago
A Simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some ty
Setler [38]

Answer: a) (0.755, 0.925)

Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

n= 100

sample proportion: \hat{p}=0.84

Significance level : \alpha= 1-0.98=0.02

Critical z-value : z_{\alpha/2}=2.33  (using z-value table)

Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

i.e. 0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}

i.e. \approx0.84\pm 0.085

i.e. (0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)

3 0
3 years ago
43. What is the translation rule to map AHIJ to AH ! J'?<br> 11<br> H
lutik1710 [3]
Left 4. Or x-4 because the way it moves
3 0
3 years ago
How do you find the equations of the vertical and horizontal asymptotes for a graph when you only have the graph and no function
-BARSIC- [3]

Answer:

See below for answer and explanations (as well as an attached graph)

Step-by-step explanation:

Pay attention to the behavior of the asymptotes. If the asymptotes are approaching a certain x-value or y-value, then that value is undefined for the function.

Take for example y=\frac{1}{x}:

  • As x approaches ∞ and -∞, then y approaches 0, which is our horizontal asymptote
  • As y approaches ∞ and -∞, then x approaches 0, which is our vertical asymptote

See the graph for a visual.

8 0
3 years ago
Someone please help me with this lol… have no idea what I’m doing
Sholpan [36]

Given:

\cos \theta =\dfrac{3}{5}

\sin \theta

To find:

The quadrant of the terminal side of \theta and find the value of \sin\theta.

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

\cos \theta =\dfrac{3}{5}

\sin \theta

Here cos is positive and sine is negative. So, \theta must be lies in Quadrant IV.

We know that,

\sin^2\theta +\cos^2\theta =1

\sin^2\theta=1-\cos^2\theta

\sin \theta=\pm \sqrt{1-\cos^2\theta}

It is only negative because \theta lies in Quadrant IV. So,

\sin \theta=-\sqrt{1-\cos^2\theta}

After substituting \cos \theta =\dfrac{3}{5}, we get

\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}

\sin \theta=-\sqrt{1-\dfrac{9}{25}}

\sin \theta=-\sqrt{\dfrac{25-9}{25}}

\sin \theta=-\sqrt{\dfrac{16}{25}}

\sin \theta=-\dfrac{4}{5}

Therefore, the correct option is B.

6 0
3 years ago
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