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3 years ago

MATH help right now plzzz HURRY NOOW NOW NOW 2 ATTACHMENTS GIVING A BRAINLIEST ANSWER ALL

Mathematics

2 answers:

Nikolay [14]3 years ago

3 0

First one is D
Second one is B.

STatiana [176]3 years ago

3 0

1 = 3rd one down.
2 = 2nd one down

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0.25:1.5 in simplest form

Lelechka [254]

The answer should be
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.16

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3 years ago

PLEASE HURRY!!!

RoseWind [281]

terms is x first then terms in y and constant after the equals

Its C

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3 years ago

Help math ITS EASY IF U LOKE RATIO

scoundrel [369]

You have the right one selected.

The answer is 2:1

To find this take each side of ABC and its corresponding side on DEF, then simplify. You can treat ratios just like fractions, so simplify them in the same way.

20:10 becomes 2:1

12:6 becomes 2:1

16:8 becomes 2:1

7 0

2 years ago

In Pennsylvania the average IQ score is 101.5. The variable is normally distributed, and the population standard deviation is 15

blsea [12.9K]

Answer:

We conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 101.5

Sample mean, $\bar{x}$ = 106.4

Sample size, n = 30

Alpha, α = 0.05

Population standard deviation, σ = 15

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 101.5\\H_A: \mu > 101.5$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789$

Now, $z_{critical} \text{ at 0.05 level of significance } = 1.96$

Since,

$z_{stat} > z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

5 0

3 years ago

in the dominican republic in august, the distribution of daily high temperature is approximately normal with mean 86 degrees fah

lilavasa [31]

The standard deviation of the distribution = 1.5°F

Given that the distribution of daily high temperature is approximately normal.

Population mean, $\mu$ = 86°F

Also approximately 95% of all daily high temperatures are between 83°F and 89°F.

So here $X_1$ = 83°F and $X_2$ = 89°F

We have the z - statistic ,

$z=\frac{x-\mu}{\sigma}$

where $\sigma$ is the Standard deviation.

For 95% probability, the z-value for normal distribution is 1.96.

As we consider the positive z-value, take X = $X_2$ = 89°F

So, $z=\frac{x-\mu}{\sigma}$

⇒ $\sigma = \frac{x-\mu}{z} = \frac{89-86}{1.96} =1.53$ ≈ 1.5°F

So the standard deviation of the distribution = 1.5°F

Learn more about standard deviation at brainly.com/question/475676

#SPJ4

5 0

1 year ago