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lord [1]
3 years ago
10

Ok pls help!! I’m stuck! And this is the last question and imma be late

Mathematics
1 answer:
enot [183]3 years ago
7 0

Answer:

64°

hahdhuehehdoefudystiuritisisiiijji

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It took Everly 55 minutes to run a 10-kilometer race last weekend. If you know that 1 kilometer equals 0.621 mile, how many minu
m_a_m_a [10]
55/6.21 = x/1....55 mins to 6.21 miles = x min to 1 mile
cross multiply
6.21x = 55
x = 55/6.21
x = 8.86 minutes
3 0
4 years ago
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The table shows the distances traveled by a paper airplane.
BabaBlast [244]

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Step-by-step explanation:

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An agronomist hopes that a new fertilizer she has developed will enable grape growers to increase the yield of each grapevine by
Phoenix [80]

Answer:

Yes. Fertilization increases grape yields by more than 5 pounds.

Step-by-step explanation:

Let f-fertilized and o-old(unfertilized)

#First, we use our data to calculate the standard error:

SE(\bar y_f-\bar y_o)=\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}\\\\=\sqrt{\frac{3.7^2}{44}+\frac{3.4^2}{47}}\\\\=0.7464

#State both null and alternative hypothesis:

H_o:\mu_f-\mu_o=0\\\\H_A=\mu_f-\mu_o>0

#We determine our degrees of freedom as 87.02(using R), we now compute the t-value as:

t=\frac{\bar y_f-\bar y_o}{\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}}\\\\\\t=\frac{53.4-52.1}{\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}}\\\\\\t=\frac{1.3}{0.7464}\\\\t=1.7417\\\\P=P(t_{87.0>1.7417}=1-0.9575=0.0.0425

Since, the p-value is low, we Reject the null hypothesis. The is enough evidence suggesting that grape yields increase by more than 5 pounds than mean yields of  unfertilized grapes.

8 0
4 years ago
What is 56/160 in simplest form? ​
denis-greek [22]

Answer:

7/20

Step-by-step explanation:

Just reduce by dividing by the GCF

5 0
3 years ago
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Write the number 0.031 in scientific notation
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Answer:

3.1 x 10^-2

Step-by-step explanation:

5 0
3 years ago
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