How do you solve the problem -x-29=13+2x

A math teacher’s midterm is worth 200 points. The mean of the test is 143 and the standard deviation is 15.7. If the scores have

Klio2033 [76]

The answer is D: about 95%.

7 0

2 years ago

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Can someone help me really please

NARA [144]

Answer:

b

Step-by-step explanation:

3/3 = 1

4 0

2 years ago

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The following polynomial can be factored using the techniques of this lesson x^2-18

Papessa [141]

Answer:

$(x+3\sqrt{2})(x-3\sqrt{2})$

Step-by-step explanation:

The given expression is $x^2-18$ .

The technique that can be used to factor this polynomial is the difference of two squares.

We rewrite the two expressions as;

$x^2-(\sqrt{18})^2$

Recall that;

$a^2-b^2=(a+b)(a-b)$

$x^2-(\sqrt{18})^2=(x+\sqrt{18})(x-\sqrt{18})$

$x^2-(\sqrt{18})^2=(x+3\sqrt{2})(x-3\sqrt{2})$

8 0

2 years ago

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At Open Skate Night, admission is $3.75 with a membership card and $5.00 without a membership card. Skate rentals are $3.00.

Rina8888 [55]

(I need more charecters) the price might be $23.75

5 0

2 years ago

A 9-year-old girl did a science fair experiment in which she tested professional touch therapists to see if they could sense he

Orlov [11]

Answer:

We conclude that the the touch therapists does not use a method equivalent to random guesses.

Yes, the results suggest that touch therapists are effective.

Step-by-step explanation:

We are given that in a 9-year-old girl did a science fair experiment in which she tested professional touch therapists to see if they could sense her energy field.

Among 264264 trials, the touch therapists were correct 105105 times.

Let p = proportion that touch therapists uses a random guess method.

Here, random guess means; p = 50%

So, Null Hypothesis, $H_0$ : p = 50% {means that the touch therapists use a method equivalent to random guesses}

Alternate Hypothesis, $H_A$ : p $\neq$ 50% {means that the touch therapists does not use a method equivalent to random guesses}

The test statistics that would be used here One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of touch therapists were correct = $\frac{105}{264}$ = 0.39

n = sample of trials = 264

So, test statistics = $\frac{\frac{105}{264} -0.50}{\sqrt{\frac{\frac{105}{264}(1-\frac{105}{264})}{264} } }$

= -3.395

The value of z test statistics is -3.395.

Now, at 0.05 significance level the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the the touch therapists does not use a method equivalent to random guesses.

Yes, the results suggest that touch therapists are effective.

3 0

2 years ago