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Anestetic [448]
3 years ago
6

WILL GIVE BRAINLIEST FOR THE CORRECT ANSWER AND POINTS!

Mathematics
1 answer:
Otrada [13]3 years ago
5 0

Answer:

c

Step-by-step explanation:

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What is the value of x?<br>ОА. 95°<br>Ов. 1050<br>Ос. 75°<br>OD. 35°​
statuscvo [17]
C. 75

70+35+x=180
70+35=105
180-105=75
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3 years ago
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Fiona has a box with 5 red buttons and 9 blue buttons.She randomly selected one button, took it out of the box, then selected an
leonid [27]

Answer:

B. 10/91

Step-by-step explanation:

Box has 5 red buttons and 9 blue buttons.

We add 5 and 9 to get the total buttons.

5 + 9 = 14

Then, we find the probability for a red button.

5/14 = .35

Now, that we've taken out 1 button.

The amount of buttons available changes.

14 - 1 = 13

There are 4 red buttons remaining.

The probability for a red button changes to 4/13.

4/13 = .30

Then, we multiply both probabilities.

.30 x .35 = .111

Convert to a fraction.

.111 = 10/91

Thus it would be B.

7 0
2 years ago
Can someone plz help me
lubasha [3.4K]
I say it’s c it makes the most sense to me
3 0
2 years ago
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4√3√2² please someone<br>​
lubasha [3.4K]

Step-by-step explanation:

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6 0
3 years ago
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The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims
kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is \mu=600, we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it \sigma) is

\sigma=48

Next they draw a random sample of n=70 students, and they got a mean score (denoted by \bar x) of \bar x=613

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis H_0:\bar x \geq \mu

- The alternative would be then the opposite H_0:\bar x < \mu

The test statistic for this type of test takes the form

t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\

<h3>since 2.266>1.645 we  can reject the null hypothesis.</h3>
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3 years ago
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