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Tasya [4]
3 years ago
5

Can someone help me please !

Mathematics
1 answer:
pickupchik [31]3 years ago
3 0
I would say C, mean because 13 text messages is much lower than the other numbers
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Let mu denote the true average number of minutes of a television commercial. Suppose the hypothesis H0: mu = 2.1 versus Ha: mu &
natka813 [3]

Answer:

A. T > 2.539

Step-by-step explanation:

We have a hypothesis test of the mean, with unknown population standard deviation.

The hypothesis are:

H_0: \mu = 2.1 \\\\H_a: \mu > 2.1

From the hypothesis we can see that the test is right-tailed, so the critical value of t should be a positive value.

The degrees of freedom can be calculated as:

df=n-1=20-1=19

The significance level is 0.01, so the critical value tc should be the one that satisfies:

P(t>t_c)=0.01

Looking up in a t-table, for 19 degrees of freedom, this critical value is tc=2.539.

7 0
3 years ago
What is value of zero in 42.02?​
Kryger [21]

tenths because 42.02 it would be tenths the first decimal is tenths and second is hundreths

7 0
3 years ago
A bag of apples weighs 7 7/8 pounds. By weight, 1/15 of the apples are rotten. What is the weight of the good apples?
siniylev [52]
The good apples are 14/15 of the total apples. So multiply the total apples by the weight of good ones to find the solution. \frac{63}{8}*\frac{14}{15} = \frac{294}{40} = 7\frac{7}{20}lbs
4 0
3 years ago
1. Identify and EXPLAIN the ERROR
Anit [1.1K]

This might be a trick question. I did the problem and got the exact same answer. I got 31 x + 0.5. If there's anything they did wrong maybe its because they didn't put a zero before 5 so 5 could become a decimal?


7 0
3 years ago
In this exercise, consider a particle moving on a circular path of radius b described by r(t) = b cos(ωt)i + b sin(ωt)j, where ω
Bingel [31]

Answer:

Acceleration of the particle = bw^{2}

Step-by-step explanation:

We are given the position vector of a particle moving in a circle of radius b units.

r(t) = b cos(ωt)i + b sin(ωt)j

Velocity , v =\frac{dr}{dt} = -bω sin(ωt)i + bω cos(ωt)j

The magnitude of velocity, v =\sqrt{v_x^{2} +v_y^{2} }

Squaring both sides,

v^{2} = b^{2} w^{2}(sin^{2}(wt)+cos^{2}(wt))

Since sin^{2}(wt)+cos^{2}(wt)) = 1

v^{2} = b^{2}w^{2}

The acceleration towards the centre is called the centripetal acceleration and is given by

a = \frac{v^{2} }{r}

a = \frac{b^{2}w^{2}}{b}

a = bw^{2}

3 0
3 years ago
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