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babymother [125]
3 years ago
12

5/32 × 64y/100 please help :,)

Mathematics
1 answer:
ladessa [460]3 years ago
3 0

Answer:

1/10y

Step-by-step explanation

(5/32(64))(y)/100

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A high school has 44 players on the football team. The summary of the players' weights is given in the box plot. Approximately,
Marina CMI [18]

Answer:

Step-by-step explanation:

The missing image is attached below.

From the given information

The number of players is 44.

From the box plot;

The minimum weight = 154 pounds

The first quartile Q₁ = 159 which is 25% of the data below 159 pounds.

The second quartile Q₂ = 213 which is 50% of the data below 213 pounds

The third quartile Q₃ = 253 which is 75% of the data below 253 pounds

The maximum weight = 268 pounds.

Here, the value of 213 is the middle value and which signifies the median.

The 50% of the data are located both on the left side and the right side of the median.

Thus, the percentage of players weighting greater than or equal to 213 pounds is 50%.

7 0
3 years ago
Https://allmylinks.com/zeoathiras<br><br> I’m bored. Owo
Savatey [412]

Answer:

Yeah same here

Step-by-step explanation:

8 0
2 years ago
I NEED HELP FAST WHAT IS 80 + -9 - 5 + 89-66+44-9+87-100+213
kow [346]

Answer:

324 is ya answer!

5 0
3 years ago
What is the answer to -3x-5&lt;-2 and what are the steps to actually finding the answer for it??
Andre45 [30]

Answer:

x > - 1

Step-by-step explanation:

Given

- 3x - 5 < - 2 ( add 5 to both sides )

- 3x < 3

Divide both sides by - 3, reversing the symbol as a result of dividing by a negative quantity

x > - 1

7 0
2 years ago
A recent broadcast of a television show had a 10 ​share, meaning that among 6000 monitored households with TV sets in​ use, 10​%
Kisachek [45]

Answer:

Null and alternative hypothesis

H_0: \pi \geq0.25\\\\H_1: \pi

Test statistic z=-26.82

P-value P=0

The null hypothesis is rejected.

It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

Step-by-step explanation:

We have to perform a hypothesis test of a proportion.

The claim is that less than 25% were tuned into the program, so we will state this null and alternative hypothesis:

H_0: \pi \geq0.25\\\\H_1: \pi

The signifiance level is 0.01.

The sample has a proportion p=0.1 and sample size of n=6000.

The standard deviation of the proportion, needed to calculate the test statistic, is:

\sigma_p=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.25(1-25)}{6000} } =0.0056

The test statistic is calculated as:

z=\frac{p-\pi+0.5/N}{\sigma_p} =\frac{0.1-0.25+0.5/6000}{0.0056}=\frac{-0.1499}{0.0056}  =-26.82

As this is a one-tailed test, the P-value is P(z<-26.82)=0. The P-value is smaller than the significance level (0.01), so the effect is significant.

Since the effect is significant, the null hypothesis is rejected. It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

3 0
3 years ago
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