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Lena [83]
3 years ago
12

A +1>3guyz at least solve one question ​

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

a  + 1 > 3

Move 1 to another side.

a > 3 - 1 \\ a > 2

Therefore, a>2

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EASY PLEASE HELP!!!!!!
Serga [27]

Answer:

e - surveys

Step-by-step explanation:

I'm not sure btw sorry but hope this can help

4 0
3 years ago
Find the slope of the line that passes through the points 2,4,6 and 12
kolezko [41]

Answer:

The slope is 2.

Step-by-step explanation:

If you use the formula that is shown in the picture, you can calculate that 12-4/6-2 is the slope. If you simplify, you'll get that 8/4 is the slope which simplifies to 2.

3 0
3 years ago
4.875 in expanded <br> form in fractions
TEA [102]
(4x10)+(8x1/10)+(7x1/100)+(5x1/1000)
6 0
3 years ago
It is believed that the average starting salary for a 21-25 year old college grad exceeds $52,000 per year. Hence, it is desired
gregori [183]

Answer:

The significance level α is 0.02.

Step-by-step explanation:

A hypothesis test for single mean can be performed to determine whether the average starting salary for a 21-25 year old college grad exceeds $52,000 per year.

The hypothesis is defined as follows:

<em>H₀</em>: The average starting salary for a 21-25 year old college grad does not exceeds $52,000 per year, i.e. <em>µ</em> ≤ 52,000.

<em>Hₐ</em>: The average starting salary for a 21-25 year old college grad exceeds $52,000 per year, i.e. <em>µ</em> > 52,000.

The information provided is:

<em>σ</em> = $5,745

<em>n</em> = 65

Also, if \bar X>\$53,460 then the null hypothesis will be rejected.

Here, we need to compute the value of significance level <em>α</em>, the type I error probability.

A type I error occurs when we reject a true null hypothesis (H<em>₀</em>).

That is:

<em>α</em> = P (type I error)

<em>α</em> = P (Rejecting H<em>₀</em>| H<em>₀</em> is true)

   =P(\bar X>53460|\mu \leq 52000)

   =P[\frac{\bar X-\mu_{0}}{\sigma/\sqrt{n}}>\frac{53460-52000}{5745/\sqrt{65}}]

   =P(Z>2.05)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the significance level α is 0.02.

5 0
4 years ago
A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le
CaHeK987 [17]

Answer:

Step-by-step explanation:

Hello!

To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.

To test this claim two samples were taken:

Sample 1

X: battery lifespan of a unit of the new product (min)

n= 93 units of the new product

mean battery life X[bar]= 8:53hs= 533min

S= 84 min

Sample 2

X: battery lifespan of a unit of the leading product (min)

n= 102 units of the leading product

mean battery life X[bar]= 5:40 hs = 340min

S= 93 min

The population variances of both variances are unknown and distinct.

To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:

H₀: μ₁ - μ₂ ≤ 120

H₁:  μ₁ - μ₂ > 120

Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:

Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}  } }

Z≈N(0;1)

Z= \frac{(533-340)-120}{\sqrt{\frac{84^2}{56} +\frac{93^2}{102}  } }= 5.028

I hope this helps!

3 0
4 years ago
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